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n!

C(n,p) = ------------

p!.(n-p)!

Properties of combinations

With the last formula it is easy to prove thata) C(n,p) = C(n,n-p)

b) C(n,p) = C(n-1,p) + C(n-1,p-1) (Pascal's formula)

Binomial coefficient

The number C(n,p) is called a binomial coefficient and this is written as

n

( )

p

Binomial theorem

We'll prove that

(a + b)n = an + C(n,1)an-1 b + C(n,2)an-2 b2 + C(n,3)an-3 b3 + ...

+ C(n,n) bn

To prove this theorem we use mathematical induction.It is easy to verify that the theorem holds for n = 2 .

Now, assume it holds for n = k. We'll show it holds for n= k+1.

(a + b)k+1 = (a + b).(a + b)k

=(a+b).(ak + C(k,1)ak-1 b + C(k,2)ak-2 b2 + C(k,3)ak-3 b3 + ... C(k,k)bk )

= ak+1 + C(k,1)ak b + C(k,2)ak-1 b2 + C(k,3)ak-2 b3 + ... C(k,k)a bk +

ak b + C(k,1)ak-1 b2 + C(k,2)ak-2 b3 + ...+ C(k,k-1)a bk + C(k,k) bk+1

Since C(k,k)= C(k+1,k+1)= 1 and appealing on Pascal's formula

C(n,p) = C(n-1,p) + C(n-1,p-1) , we find

= ak+1 + C(k+1,1)ak b + C(k+1,2)ak-1 b2 + C(k+1,3)ak-2 b3 + ...

+ C(k+1,k+1) bk+1

This proves the theorem.

Counting with repetition

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Variations with repetition

Take a set A of n different elements. Point p elements in a specific order.Each such choice is called a variation with repetition of n elements choose p. Well, there are n

possibilities to point each element.

The total number of variations with repetition of n elements choose p is (np

).

We write

V'(n,p) = np

Permutations with repetition

Example :Take 3 red marbles, 2 blue marbles, and 5 yellow ones. Place this marbles in a specific order.

We call this a permutations with repetition of the marbles. Now, we'll calculate the number ofsuch permutations.

Take 10 numbered compartments. First place the three red marbles in a compartment. This givesC(10,3) possibilities. Then place the 2 blue marbles. Now there are C(10-3,2) possibilities. At

last, we place the 5 yellow marbles. Now there are C(10-3-2,5) possibilities.

The total number of possibilities are

C(10,3).C(10-3,2).C(10-3-2,5)

10! 7! 5!

= -------.-------. ------

3!.7! 2!.5! 5!.0!

10!

= --------------

3! . 2! . 5!

This method can easily be generalized.

Combinations with repetition

Take n different and ordered elements.

Point p elements one after another and order these elements in the same order as the givenelements.

The result is called a combination with repetition of n elements choose p.

We write the number of all combinations with repetition of n elements choose p as C'(n,p).

Example: A = (a,b,c,d,e) and p = 6

Then (a, a, b, d, d, d) ; (b, b, b, c, d, e) ; (c, c, c, c, c, c)are combinations with repetition of 5 elements choose 6.

We can represent such combination by means of a symbol with points and slashes.

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(a,a,b,d,d,d) .././/.../

(b,b,b,c,d,e) /.../././.

(c;c;c;c;c;c) //......//

Each symbol consists of 10 places with exactly 6 points and four slashes. With each such

combination there is just one symbol and with each symbol corresponds just one such

combination.

We can build a symbol putting exactly 6 points in 10 places.

Afterwards, the spare places are filled with slashes.

This can be done in C(10,6) ways.

So, C'(5,6) = C(5+6-1,6)

Generalizing you can prove that C'(n,p) = C(n+p-1,p)