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Combinatorics Pascal’s 4 Proofs ???
Combinatorial Enumeration in Pascal’s TriangleHow To Count Without Counting
Brian K. Miceli
Trinity UniversityMathematics Department
Mathematics Majors’ SeminarMarch 23, 2016
Miceli Combinatorics
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Combinatorics Pascal’s 4 Proofs ???
Outline
1 What is Combinatorics?
2 Pascal’s Triangle
3 Proofs
4 One Last Thing
Miceli Combinatorics
![Page 3: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/3.jpg)
Combinatorics Pascal’s 4 Proofs ???
Outline
1 What is Combinatorics?
2 Pascal’s Triangle
3 Proofs
4 One Last Thing
Miceli Combinatorics
![Page 4: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/4.jpg)
Combinatorics Pascal’s 4 Proofs ???
Outline
1 What is Combinatorics?
2 Pascal’s Triangle
3 Proofs
4 One Last Thing
Miceli Combinatorics
![Page 5: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/5.jpg)
Combinatorics Pascal’s 4 Proofs ???
Outline
1 What is Combinatorics?
2 Pascal’s Triangle
3 Proofs
4 One Last Thing
Miceli Combinatorics
![Page 6: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/6.jpg)
Combinatorics Pascal’s 4 Proofs ???
What is Combinatorics?
Combinatorics (MATH 3343) is a class offered in the fall of 2016.
As a branch of mathematics, I like to say that it’s the Art ofCounting, and it has its own style of proof, called (unsurprisingly)the combinatorial proof.
This is best illustrated with an example.
Miceli Combinatorics
![Page 7: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/7.jpg)
Combinatorics Pascal’s 4 Proofs ???
What is Combinatorics?
Combinatorics (MATH 3343) is a class offered in the fall of 2016.
As a branch of mathematics, I like to say that it’s the Art ofCounting, and it has its own style of proof, called (unsurprisingly)the combinatorial proof.
This is best illustrated with an example.
Miceli Combinatorics
![Page 8: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/8.jpg)
Combinatorics Pascal’s 4 Proofs ???
What is Combinatorics?
Combinatorics (MATH 3343) is a class offered in the fall of 2016.
As a branch of mathematics, I like to say that it’s the Art ofCounting, and it has its own style of proof, called (unsurprisingly)the combinatorial proof.
This is best illustrated with an example.
Miceli Combinatorics
![Page 9: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/9.jpg)
Combinatorics Pascal’s 4 Proofs ???
What is Combinatorics?
Combinatorics (MATH 3343) is a class offered in the fall of 2016.
As a branch of mathematics, I like to say that it’s the Art ofCounting, and it has its own style of proof, called (unsurprisingly)the combinatorial proof.
This is best illustrated with an example.
Miceli Combinatorics
![Page 10: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/10.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this. But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2. Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
![Page 11: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/11.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this. But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2. Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
![Page 12: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/12.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?
(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this. But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2. Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
![Page 13: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/13.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this.
But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2. Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
![Page 14: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/14.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this. But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2.
Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
![Page 15: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/15.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this. But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2. Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
![Page 16: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/16.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial ProblemThe proof continued . . .
(ii) Now organize the winning pairs by who is alphabetically first.That is, the number of winning pairs is
#(A is 1st) + #(B is 1st) + · · ·+ #(N is 1st) + #(O is 1st),
which isn + (n − 1) + · · ·+ 1 + 0,
which completes the proof.
Miceli Combinatorics
![Page 17: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/17.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial ProblemThe proof continued . . .
(ii) Now organize the winning pairs by who is alphabetically first.
That is, the number of winning pairs is
#(A is 1st) + #(B is 1st) + · · ·+ #(N is 1st) + #(O is 1st),
which isn + (n − 1) + · · ·+ 1 + 0,
which completes the proof.
Miceli Combinatorics
![Page 18: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/18.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial ProblemThe proof continued . . .
(ii) Now organize the winning pairs by who is alphabetically first.That is, the number of winning pairs is
#(A is 1st) + #(B is 1st) + · · ·+ #(N is 1st) + #(O is 1st),
which isn + (n − 1) + · · ·+ 1 + 0,
which completes the proof.
Miceli Combinatorics
![Page 19: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/19.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial ProblemThe proof continued . . .
(ii) Now organize the winning pairs by who is alphabetically first.That is, the number of winning pairs is
#(A is 1st) + #(B is 1st) + · · ·+ #(N is 1st) + #(O is 1st),
which isn + (n − 1) + · · ·+ 1 + 0,
which completes the proof.
Miceli Combinatorics
![Page 20: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/20.jpg)
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsDefinitions
We define
(n
k
)=
n!
k!(n − k)!=
n(n − 1) · · · (n − k + 1)
k!,
where n! = n(n − 1) · · · (2)(1).
Then n! corresponds to the number of ways of ordering n distinctobjects.
We then have that(nk
)corresponds to the number of k-element
subsets of an n-element set. By this definition, we also let(nk
)= 0
if n < 0, k < 0, or k > n.
Miceli Combinatorics
![Page 21: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/21.jpg)
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsDefinitions
We define
(n
k
)=
n!
k!(n − k)!=
n(n − 1) · · · (n − k + 1)
k!,
where n! = n(n − 1) · · · (2)(1).
Then n! corresponds to the number of ways of ordering n distinctobjects.
We then have that(nk
)corresponds to the number of k-element
subsets of an n-element set. By this definition, we also let(nk
)= 0
if n < 0, k < 0, or k > n.
Miceli Combinatorics
![Page 22: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/22.jpg)
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsDefinitions
We define
(n
k
)=
n!
k!(n − k)!=
n(n − 1) · · · (n − k + 1)
k!,
where n! = n(n − 1) · · · (2)(1).
Then n! corresponds to the number of ways of ordering n distinctobjects.
We then have that(nk
)corresponds to the number of k-element
subsets of an n-element set. By this definition, we also let(nk
)= 0
if n < 0, k < 0, or k > n.
Miceli Combinatorics
![Page 23: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/23.jpg)
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsDefinitions
We define
(n
k
)=
n!
k!(n − k)!=
n(n − 1) · · · (n − k + 1)
k!,
where n! = n(n − 1) · · · (2)(1).
Then n! corresponds to the number of ways of ordering n distinctobjects.
We then have that(nk
)corresponds to the number of k-element
subsets of an n-element set.
By this definition, we also let(nk
)= 0
if n < 0, k < 0, or k > n.
Miceli Combinatorics
![Page 24: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/24.jpg)
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsDefinitions
We define
(n
k
)=
n!
k!(n − k)!=
n(n − 1) · · · (n − k + 1)
k!,
where n! = n(n − 1) · · · (2)(1).
Then n! corresponds to the number of ways of ordering n distinctobjects.
We then have that(nk
)corresponds to the number of k-element
subsets of an n-element set. By this definition, we also let(nk
)= 0
if n < 0, k < 0, or k > n.
Miceli Combinatorics
![Page 25: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/25.jpg)
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsExamples
3! = 3 · 2 · 1 = 6, so there should be six ways of ordering theelements of the set {a, b, c}:
abc, acb, bac, bca, cab, cba.
(4
2
)=
4!
2!(4− 2)!= 6, so there should be six 2-element subsets of
the set {A,B,C ,D}:
{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.
Miceli Combinatorics
![Page 26: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/26.jpg)
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsExamples
3! = 3 · 2 · 1 = 6, so there should be six ways of ordering theelements of the set {a, b, c}:
abc, acb, bac, bca, cab, cba.
(4
2
)=
4!
2!(4− 2)!= 6, so there should be six 2-element subsets of
the set {A,B,C ,D}:
{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.
Miceli Combinatorics
![Page 27: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/27.jpg)
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsExamples
3! = 3 · 2 · 1 = 6, so there should be six ways of ordering theelements of the set {a, b, c}:
abc, acb, bac, bca, cab, cba.
(4
2
)=
4!
2!(4− 2)!= 6, so there should be six 2-element subsets of
the set {A,B,C ,D}:
{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.
Miceli Combinatorics
![Page 28: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/28.jpg)
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsExamples
3! = 3 · 2 · 1 = 6, so there should be six ways of ordering theelements of the set {a, b, c}:
abc, acb, bac, bca, cab, cba.
(4
2
)=
4!
2!(4− 2)!= 6, so there should be six 2-element subsets of
the set {A,B,C ,D}:
{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.
Miceli Combinatorics
![Page 29: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/29.jpg)
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsExamples
3! = 3 · 2 · 1 = 6, so there should be six ways of ordering theelements of the set {a, b, c}:
abc, acb, bac, bca, cab, cba.
(4
2
)=
4!
2!(4− 2)!= 6, so there should be six 2-element subsets of
the set {A,B,C ,D}:
{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.
Miceli Combinatorics
![Page 30: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/30.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Miceli Combinatorics
![Page 31: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/31.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (An algebraic proof.)(n − 1
k
)+
(n − 1
k − 1
)=
(n − 1)!
k!(n − 1− k)!+
(n − 1)!
(k − 1)!(n − k)!
=(n − k)(n − 1)!
k!(n − 1− k)!(n − k)+
k(n − 1)!
k(k − 1)!(n − k)!
=(n − k)(n − 1)! + k(n − 1)!
k!(n − k)!
=n(n − 1)!
k!(n − k)!=
(n
k
)
Miceli Combinatorics
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Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (A combinatorial proof.) How many ways are there tochoose a k-element subset from the set {1, 2, . . . n}?(i) By definition, there are
(nk
)ways to do this.
(ii) Alternatively, we can divide all such subsets into two cases bywhether or not our k-element subsets contain n. If n is not in oursubset, then we must choose our k elements from the set{1, 2, . . . , n − 1}, and there are
(n−1k
)ways to do that. If n is in
our subset, then we must choose the other k − 1 elements of oursubset from the set {1, 2, . . . , n − 1}, and there are
(n−1k−1
)ways to
do that.
Miceli Combinatorics
![Page 33: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/33.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (A combinatorial proof.) How many ways are there tochoose a k-element subset from the set {1, 2, . . . n}?(i) By definition, there are
(nk
)ways to do this.
(ii) Alternatively, we can divide all such subsets into two cases bywhether or not our k-element subsets contain n. f n is not in oursubset, then we must choose our k elements from the set{1, 2, . . . , n − 1}, and there are
(n−1k
)ways to do that. If n is in
our subset, then we must choose the other k − 1 elements of oursubset from the set {1, 2, . . . , n − 1}, and there are
(n−1k−1
)ways to
do that.
Miceli Combinatorics
![Page 34: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/34.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (A combinatorial proof.) How many ways are there tochoose a k-element subset from the set {1, 2, . . . n}?(i) By definition, there are
(nk
)ways to do this.
(ii) Alternatively, we can divide all such subsets into two cases bywhether or not our k-element subsets contain n. If n is not in oursubset, then we must choose our k elements from the set{1, 2, . . . , n − 1}, and there are
(n−1k
)ways to do that. If n is in
our subset, then we must choose the other k − 1 elements of oursubset from the set {1, 2, . . . , n − 1}, and there are
(n−1k−1
)ways to
do that.
Miceli Combinatorics
![Page 35: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/35.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (A combinatorial proof.) How many ways are there tochoose a k-element subset from the set {1, 2, . . . n}?(i) By definition, there are
(nk
)ways to do this.
(ii) Alternatively, we can divide all such subsets into two cases bywhether or not our k-element subsets contain n. If n is not in oursubset, then we must choose our k elements from the set{1, 2, . . . , n − 1}, and there are
(n−1k
)ways to do that. If n is in
our subset, then we must choose the other k − 1 elements of oursubset from the set {1, 2, . . . , n − 1}, and there are
(n−1k−1
)ways to
do that.
Miceli Combinatorics
![Page 36: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/36.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (A combinatorial proof.) How many ways are there tochoose a k-element subset from the set {1, 2, . . . n}?(i) By definition, there are
(nk
)ways to do this.
(ii) Alternatively, we can divide all such subsets into two cases bywhether or not our k-element subsets contain n. If n is not in oursubset, then we must choose our k elements from the set{1, 2, . . . , n − 1}, and there are
(n−1k
)ways to do that. If n is in
our subset, then we must choose the other k − 1 elements of oursubset from the set {1, 2, . . . , n − 1}, and there are
(n−1k−1
)ways to
do that.
Miceli Combinatorics
![Page 37: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/37.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Pascal’s Triangle is named after Blaise Pascal, a Frenchmathematician in the 1600’s.
Pascal’s Triangle was known in China as early as the late 1200’s.
Miceli Combinatorics
![Page 38: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/38.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Pascal’s Triangle is named after Blaise Pascal, a Frenchmathematician in the 1600’s.
Pascal’s Triangle was known in China as early as the late 1200’s.
Miceli Combinatorics
![Page 39: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/39.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Pascal’s Triangle is named after Blaise Pascal, a Frenchmathematician in the 1600’s.
Pascal’s Triangle was known in China as early as the late 1200’s.
Miceli Combinatorics
![Page 40: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/40.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Pascal’s Triangle is named after Blaise Pascal, a Frenchmathematician in the 1600’s.
Pascal’s Triangle was known in China as early as the late 1200’s.
Miceli Combinatorics
![Page 41: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/41.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Miceli Combinatorics
![Page 42: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/42.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
Miceli Combinatorics
![Page 43: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/43.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
Each entry is the sum of two numbers: the one directly above andthe one to the left of that one, where we assume any blank spaceis a 0.
Miceli Combinatorics
![Page 44: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/44.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
We refer to the element in the nth row and kth column as Pn,k .
Miceli Combinatorics
![Page 45: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/45.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
We refer to the element in the nth row and kth column as Pn,k .For example, P3,2 = 3,
Miceli Combinatorics
![Page 46: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/46.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
We refer to the element in the nth row and kth column as Pn,k .For example, P3,2 = 3, P4,0 = 1,
Miceli Combinatorics
![Page 47: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/47.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
We refer to the element in the nth row and kth column as Pn,k .For example, P3,2 = 3, P4,0 = 1, P5,3 = 10,
Miceli Combinatorics
![Page 48: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/48.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
We refer to the element in the nth row and kth column as Pn,k .For example, P3,2 = 3, P4,0 = 1, P5,3 = 10, P6,9 = 0.
Miceli Combinatorics
![Page 49: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/49.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
By the known relation in Pascal’s Triangle,Pn,k = Pn−1,k + Pn−1,k−1 for any n, k ∈ Z.
Miceli Combinatorics
![Page 50: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/50.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Theorem
For every n, k ∈ Z,(n
k
)= Pn,k .
Proof: As stated, Pn,k = Pn−1,k + Pn−1,k−1, with Pn,k = 0 ifn < 0, k < 0, or n < k , and from Pascal’s Triangle, P0,0 = 1.
We can compute that(00
)= 1, and we have that
(nk
)= 0 whenever
n < 0, k < 0, or n < k .
Since(nk
)=(n−1
k
)+(n−1k−1
), we see that Pn,k and
(nk
)satisfy the
same recursions and initial conditions, so it must be the case that(nk
)= Pn,k for every n, k ∈ Z.
Miceli Combinatorics
![Page 51: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/51.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Theorem
For every n, k ∈ Z,(n
k
)= Pn,k .
Proof: As stated, Pn,k = Pn−1,k + Pn−1,k−1, with Pn,k = 0 ifn < 0, k < 0, or n < k , and from Pascal’s Triangle, P0,0 = 1.
We can compute that(00
)= 1, and we have that
(nk
)= 0 whenever
n < 0, k < 0, or n < k .
Since(nk
)=(n−1
k
)+(n−1k−1
), we see that Pn,k and
(nk
)satisfy the
same recursions and initial conditions, so it must be the case that(nk
)= Pn,k for every n, k ∈ Z.
Miceli Combinatorics
![Page 52: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/52.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Theorem
For every n, k ∈ Z,(n
k
)= Pn,k .
Proof: As stated, Pn,k = Pn−1,k + Pn−1,k−1, with Pn,k = 0 ifn < 0, k < 0, or n < k , and from Pascal’s Triangle, P0,0 = 1.
We can compute that(00
)= 1, and we have that
(nk
)= 0 whenever
n < 0, k < 0, or n < k .
Since(nk
)=(n−1
k
)+(n−1k−1
), we see that Pn,k and
(nk
)satisfy the
same recursions and initial conditions, so it must be the case that(nk
)= Pn,k for every n, k ∈ Z.
Miceli Combinatorics
![Page 53: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/53.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Theorem
For every n, k ∈ Z,(n
k
)= Pn,k .
Proof: As stated, Pn,k = Pn−1,k + Pn−1,k−1, with Pn,k = 0 ifn < 0, k < 0, or n < k , and from Pascal’s Triangle, P0,0 = 1.
We can compute that(00
)= 1, and we have that
(nk
)= 0 whenever
n < 0, k < 0, or n < k .
Since(nk
)=(n−1
k
)+(n−1k−1
), we see that Pn,k and
(nk
)satisfy the
same recursions and initial conditions, so it must be the case that(nk
)= Pn,k for every n, k ∈ Z.
Miceli Combinatorics
![Page 54: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/54.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Theorem
For every n, k ∈ Z,(n
k
)= Pn,k .
Proof: As stated, Pn,k = Pn−1,k + Pn−1,k−1, with Pn,k = 0 ifn < 0, k < 0, or n < k , and from Pascal’s Triangle, P0,0 = 1.
We can compute that(00
)= 1, and we have that
(nk
)= 0 whenever
n < 0, k < 0, or n < k .
Since(nk
)=(n−1
k
)+(n−1k−1
), we see that Pn,k and
(nk
)satisfy the
same recursions and initial conditions, so it must be the case that(nk
)= Pn,k for every n, k ∈ Z.
Miceli Combinatorics
![Page 55: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/55.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
In other words, Pascal’s Triangle looks like this:(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)(60
) (61
) (62
) (63
) (64
) (65
) (66
)Keeping this in mind, let’s see what we can conjecture about thenumbers in Pascal’s Triangle . . .
Miceli Combinatorics
![Page 56: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/56.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
In other words, Pascal’s Triangle looks like this:
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)(60
) (61
) (62
) (63
) (64
) (65
) (66
)Keeping this in mind, let’s see what we can conjecture about thenumbers in Pascal’s Triangle . . .
Miceli Combinatorics
![Page 57: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/57.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
In other words, Pascal’s Triangle looks like this:(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)(60
) (61
) (62
) (63
) (64
) (65
) (66
)
Keeping this in mind, let’s see what we can conjecture about thenumbers in Pascal’s Triangle . . .
Miceli Combinatorics
![Page 58: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/58.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
In other words, Pascal’s Triangle looks like this:(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)(60
) (61
) (62
) (63
) (64
) (65
) (66
)Keeping this in mind, let’s see what we can conjecture about thenumbers in Pascal’s Triangle . . .
Miceli Combinatorics
![Page 59: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/59.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
Miceli Combinatorics
![Page 60: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/60.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get when sum across the rows of Pascal’s Triangle?
Miceli Combinatorics
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Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum from the tops of the columns and godown to some stopping point?
Miceli Combinatorics
![Page 62: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/62.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we take an alternating sum across the rows?
Miceli Combinatorics
![Page 63: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/63.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum diagonally, starting at the far right andworking down and left?
Miceli Combinatorics
![Page 64: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/64.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
Miceli Combinatorics
![Page 65: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/65.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get when sum across the rows of Pascal’s Triangle?
n∑k=0
(n
k
)= 2n
Miceli Combinatorics
![Page 66: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/66.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get when sum across the rows of Pascal’s Triangle?
n∑k=0
(n
k
)= 2n
Miceli Combinatorics
![Page 67: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/67.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum from the tops of the columns and godown to some stopping point?
n∑i=k
(i
k
)=
(n + 1
k + 1
)
Miceli Combinatorics
![Page 68: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/68.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum from the tops of the columns and godown to some stopping point?
n∑i=k
(i
k
)=
(n + 1
k + 1
)
Miceli Combinatorics
![Page 69: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/69.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we take an alternating sum across the rows?
n∑k=0
(−1)k(n
k
)= 0
Miceli Combinatorics
![Page 70: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/70.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we take an alternating sum across the rows?
n∑k=0
(−1)k(n
k
)= 0
Miceli Combinatorics
![Page 71: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/71.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum diagonally, starting at the far right andworking down and left? ∑
i+j=n,i≤j
(j
i
)= fn
Miceli Combinatorics
![Page 72: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/72.jpg)
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum diagonally, starting at the far right andworking down and left? ∑
i+j=n,i≤j
(j
i
)= fn
Miceli Combinatorics
![Page 73: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/73.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the First Conjecture
Theorem
For n ≥ 0,n∑
k=0
(n
k
)= 2n.
Proof: How many subsets of A = {1, 2, . . . , n} exist?
(i) From Intro to Abstract, we know there are 2n subsets of A.
(ii) The total number of subsets is also equal to
#(0-elmt. subsets)+#(1-elmt. subsets)+· · ·+#(n-elmt. subsets),
which is (n
0
)+
(n
1
)+ · · ·+
(n
n
).
Miceli Combinatorics
![Page 74: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/74.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the First Conjecture
Theorem
For n ≥ 0,n∑
k=0
(n
k
)= 2n.
Proof: How many subsets of A = {1, 2, . . . , n} exist?
(i) From Intro to Abstract, we know there are 2n subsets of A.
(ii) The total number of subsets is also equal to
#(0-elmt. subsets)+#(1-elmt. subsets)+· · ·+#(n-elmt. subsets),
which is (n
0
)+
(n
1
)+ · · ·+
(n
n
).
Miceli Combinatorics
![Page 75: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/75.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the First Conjecture
Theorem
For n ≥ 0,n∑
k=0
(n
k
)= 2n.
Proof: How many subsets of A = {1, 2, . . . , n} exist?
(i) From Intro to Abstract, we know there are 2n subsets of A.
(ii) The total number of subsets is also equal to
#(0-elmt. subsets)+#(1-elmt. subsets)+· · ·+#(n-elmt. subsets),
which is (n
0
)+
(n
1
)+ · · ·+
(n
n
).
Miceli Combinatorics
![Page 76: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/76.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the First Conjecture
Theorem
For n ≥ 0,n∑
k=0
(n
k
)= 2n.
Proof: How many subsets of A = {1, 2, . . . , n} exist?
(i) From Intro to Abstract, we know there are 2n subsets of A.
(ii) The total number of subsets is also equal to
#(0-elmt. subsets)+#(1-elmt. subsets)+· · ·+#(n-elmt. subsets),
which is (n
0
)+
(n
1
)+ · · ·+
(n
n
).
Miceli Combinatorics
![Page 77: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/77.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Theorem
For n ≥ 1,n∑
k=0
(−1)k(n
k
)= 0.
Before we begin the proof, let’s think about what this theorem isasking us to show.
In terms of numbers of subsets of A = {1, 2, . . . , n}, we want toshow
#(even subsets of A)−#(odd subsets of A) = 0,
or#(even subsets of A) = #(odd subsets of A).
Miceli Combinatorics
![Page 78: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/78.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Theorem
For n ≥ 1,n∑
k=0
(−1)k(n
k
)= 0.
Before we begin the proof, let’s think about what this theorem isasking us to show.
In terms of numbers of subsets of A = {1, 2, . . . , n}, we want toshow
#(even subsets of A)−#(odd subsets of A) = 0,
or#(even subsets of A) = #(odd subsets of A).
Miceli Combinatorics
![Page 79: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/79.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Theorem
For n ≥ 1,n∑
k=0
(−1)k(n
k
)= 0.
Before we begin the proof, let’s think about what this theorem isasking us to show.
In terms of numbers of subsets of A = {1, 2, . . . , n}, we want toshow
#(even subsets of A)−#(odd subsets of A) = 0,
or#(even subsets of A) = #(odd subsets of A).
Miceli Combinatorics
![Page 80: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/80.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Theorem
For n ≥ 1,n∑
k=0
(−1)k(n
k
)= 0.
Before we begin the proof, let’s think about what this theorem isasking us to show.
In terms of numbers of subsets of A = {1, 2, . . . , n}, we want toshow
#(even subsets of A)−#(odd subsets of A) = 0,
or#(even subsets of A) = #(odd subsets of A).
Miceli Combinatorics
![Page 81: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/81.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Theorem
For n ≥ 1,n∑
k=0
(−1)k(n
k
)= 0.
Before we begin the proof, let’s think about what this theorem isasking us to show.
In terms of numbers of subsets of A = {1, 2, . . . , n}, we want toshow
#(even subsets of A)−#(odd subsets of A) = 0,
or#(even subsets of A) = #(odd subsets of A).
Miceli Combinatorics
![Page 82: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/82.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Proof: Let En denote the even subsets of A, and let On denote theodd subset of A.
Define f : En → On by
i. f (X ) = X ∪ {1} if 1 /∈ X , and
ii. f (X ) = X − {1} if 1 ∈ X .
Then f is invertible, so |En| = |On|.
Miceli Combinatorics
![Page 83: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/83.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Proof: Let En denote the even subsets of A, and let On denote theodd subset of A.
Define f : En → On by
i. f (X ) = X ∪ {1} if 1 /∈ X , and
ii. f (X ) = X − {1} if 1 ∈ X .
Then f is invertible, so |En| = |On|.
Miceli Combinatorics
![Page 84: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/84.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Proof: Let En denote the even subsets of A, and let On denote theodd subset of A.
Define f : En → On by
i. f (X ) = X ∪ {1} if 1 /∈ X , and
ii. f (X ) = X − {1} if 1 ∈ X .
Then f is invertible, so |En| = |On|.
Miceli Combinatorics
![Page 85: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/85.jpg)
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Proof: Let En denote the even subsets of A, and let On denote theodd subset of A.
Define f : En → On by
i. f (X ) = X ∪ {1} if 1 /∈ X , and
ii. f (X ) = X − {1} if 1 ∈ X .
Then f is invertible, so |En| = |On|.
Miceli Combinatorics
![Page 86: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/86.jpg)
Combinatorics Pascal’s 4 Proofs ???
Proofs of the Other Conjectures
Theorem
For n ≥ 0,n∑
i=k
(i
k
)=
(n + 1
k + 1
).
Theorem
For n ≥ 0,∑
i+j=n,i≤j
(j
i
)= fn.
Proof: Take 3343 in the fall.
Miceli Combinatorics
![Page 87: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/87.jpg)
Combinatorics Pascal’s 4 Proofs ???
Proofs of the Other Conjectures
Theorem
For n ≥ 0,n∑
i=k
(i
k
)=
(n + 1
k + 1
).
Theorem
For n ≥ 0,∑
i+j=n,i≤j
(j
i
)= fn.
Proof: Take 3343 in the fall.
Miceli Combinatorics
![Page 88: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/88.jpg)
Combinatorics Pascal’s 4 Proofs ???
Proofs of the Other Conjectures
Theorem
For n ≥ 0,n∑
i=k
(i
k
)=
(n + 1
k + 1
).
Theorem
For n ≥ 0,∑
i+j=n,i≤j
(j
i
)= fn.
Proof:
Take 3343 in the fall.
Miceli Combinatorics
![Page 89: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/89.jpg)
Combinatorics Pascal’s 4 Proofs ???
Proofs of the Other Conjectures
Theorem
For n ≥ 0,n∑
i=k
(i
k
)=
(n + 1
k + 1
).
Theorem
For n ≥ 0,∑
i+j=n,i≤j
(j
i
)= fn.
Proof: Take 3343 in the fall.
Miceli Combinatorics
![Page 90: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/90.jpg)
Combinatorics Pascal’s 4 Proofs ???
Something to Recall from 1312
Let g(x) =∑n≥0
xn = 1 + x + x2 + x3 + · · · .
In Calc II we learn that g(x) =1
1− x, provided that |x | < 1.
One way to think of this is that g(x) is the Maclaurin series for1
1−x , which has an interval of convergence of (−1, 1).
Miceli Combinatorics
![Page 91: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/91.jpg)
Combinatorics Pascal’s 4 Proofs ???
Something to Recall from 1312
Let g(x) =∑n≥0
xn = 1 + x + x2 + x3 + · · · .
In Calc II we learn that g(x) =1
1− x, provided that |x | < 1.
One way to think of this is that g(x) is the Maclaurin series for1
1−x , which has an interval of convergence of (−1, 1).
Miceli Combinatorics
![Page 92: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/92.jpg)
Combinatorics Pascal’s 4 Proofs ???
Something to Recall from 1312
Let g(x) =∑n≥0
xn = 1 + x + x2 + x3 + · · · .
In Calc II we learn that g(x) =1
1− x, provided that |x | < 1.
One way to think of this is that g(x) is the Maclaurin series for1
1−x , which has an interval of convergence of (−1, 1).
Miceli Combinatorics
![Page 93: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/93.jpg)
Combinatorics Pascal’s 4 Proofs ???
Something to Recall from 1312
Let g(x) =∑n≥0
xn = 1 + x + x2 + x3 + · · · .
In Calc II we learn that g(x) =1
1− x, provided that |x | < 1.
One way to think of this is that g(x) is the Maclaurin series for1
1−x , which has an interval of convergence of (−1, 1).
Miceli Combinatorics
![Page 94: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/94.jpg)
Combinatorics Pascal’s 4 Proofs ???
Why Does This Happen?
What is the Maclaurin series for F (x) =1
1− x − x2?
F (x) = 1 + x + 2x2 + 3x3 + 5x4 + 8x5 + 13x6 + 21x7 + 34x8 + · · ·=
∑n≥0
fnxn.
Miceli Combinatorics
![Page 95: Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count](https://reader035.fdocuments.net/reader035/viewer/2022081402/5f159b5c7c2f5a0f185474cd/html5/thumbnails/95.jpg)
Combinatorics Pascal’s 4 Proofs ???
Why Does This Happen?
What is the Maclaurin series for F (x) =1
1− x − x2?
F (x) = 1 + x + 2x2 + 3x3 + 5x4 + 8x5 + 13x6 + 21x7 + 34x8 + · · ·
=∑n≥0
fnxn.
Miceli Combinatorics
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Combinatorics Pascal’s 4 Proofs ???
Why Does This Happen?
What is the Maclaurin series for F (x) =1
1− x − x2?
F (x) = 1 + x + 2x2 + 3x3 + 5x4 + 8x5 + 13x6 + 21x7 + 34x8 + · · ·=
∑n≥0
fnxn.
Miceli Combinatorics
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Combinatorics Pascal’s 4 Proofs ???
The End
Thanks for listening.
Questions?
Miceli Combinatorics
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Combinatorics Pascal’s 4 Proofs ???
The End
Thanks for listening.Questions?
Miceli Combinatorics