59

After learning this chapter you will be able to* Understand the concept of permutations and combinations.

* Distinguish between permutation and combination.

* Understand factorial notation.

* Understand the meanings of nPr and nC

r.

* Develop the skill in solving the problems.

Introduction :In our daily life we come across many situations wherein selection and arrangements

of objects are necessary.

A boy wants to buy 3 different books from a Book shop. In that shop thereare 5 different books. He selects 3 books out of 5. Selection of 3 books out of 5is a combination. He arranges these books in his shelf in different ways everyday. Eachway of the arrangement of books is a permutation.

Permutation is an orderly arrangement of objects

Combination is mere selection of objects

Observe :The concept of permutations and combinations have applications insolving problems of probability and genetic engineering.

PERMUTATIONS

Consider the following examples

Example 1 : There are three boys : ASHRITH (A), BHARATH (B) and CHETAN (C).

Arrange them in a row, two at a time.

The arrangements are

(i) A B (ii) A C

(iii) B A (iv) B C

(v) C A (vi) C B

There are 6 different ways of arrangements.

These arrangements are called permutations.

Think :

Can there be any morearrangement?

Is AB different from BA?

2 PERMUTATIONS AND COMBINATIONS

60

Example 2 : You are given three different books. Physics (P), Chemistry (C) andMathematics (M).

Arrange them in a row taking all the three at a time.

The arrangements are(i) P C M (ii) P M C

(iii) C P M (iv) C M P

(v) M P C (vi) M C P

Out of three books, taken all at a time there are six permutations.

Example 3 : There are four different digits 2, 3, 4, 5. How many numbers can beformed using two digits at a time?

There are 12 numbers. From four different digits, twelve 2-digit numbers areformed. There are 12 permutations.

The results of the above three examples are as follows:-Ex : No. Total no of Objects taken Symbol (nP

r) No of

objects (n) at a time (r) permutations

Example (1) 3 2 3P2

6

Example (2) 3 3 3P3

6

Example (3) 4 2 4P2

12

(i) Permutations of 3 objects taken 2 at a time is written as 3P2. This will be readas permutations of 3 objects taking two at a time.

(ii) Permutations of 3 objects taken 3 at a time is 3P3.

(iii) Permutations of 4 objects taken 2 at a time is 4P2. Hence permutations of “n” objectstaking “r” at a time is denoted by nPr.

Remember : In nPr, n and r are positive integers. rn ≥

nPr means the number of permutations of “n” things taking “r” things at

a timeAn act of arrangement of objects in an orderly way is permutation

2

3 54

23 24

25

3

2 54

32 34

35

4

2 53

42 43

45

5

2 43

52 53

54

61

1. Fundamental Principle of Counting :Example 1 : In a cinema theatre there are three entrance doors A, B, C and Two

different exit doors P and Q.

(i) There are two ways of leaving the theatre if he entersthrough A. They are AP and AQ.

(ii) If he enters through the door B then the ways of leavingthe theatre are BP and BQ.

(iii) If he enters through the door C then the exit ways areCP and CQ.

Altogether there are 2 + 2 + 2 = 6 different ways.

For three different entrance doors and two different exit doors we have3 x 2 = 6 different ways of entering and leaving the theatre.

Example 2 : There are 3 towns A, B,C. There are 3 different roads from A to Band 4 different roads from town B to C.

In How many ways can a person go from A to C through B?

A to B B to C Total

3 ways 4 ways 3 x 4 = 12 ways

xa, xb, xc, xd, ya, yb, yc, yd, za, zb, zc, zd = 12 ways

A

B

C

Q

P

A

B

C

Q

P

A

B

C

Q

P

A

x

y

z B

a

b

c

d

C

62

If one event occurs in ‘m’ different ways and another event occursindependently in ‘n’ different ways then the two events together canbe done in (m x n) different ways.

Worked Example :How many different outfits are possible if a boy has 3 shirts and 2 pants?

Solution : The boy has two choices of a pant for each of the 3 shirts.

He can wear 3 shirts in 3 different ways.

He can wear 2 pants in 2 different ways.

By Fundamental Principle, the number of ways = 3 x 2 = 6

Verification : Let us name the shirts

S1,S

2 and S

3

Pants P1 and P

2

The different outfits are

S1P

1, S

1P

2, S

2P

1, S

2P

2, S

3P

1 and S

3P

2 = 6.

2. To find the formula for number of Permutations of ‘n’ things taken ‘r’ at a time

There are n different objects and r blank boxes in a row.

Filling of r blank boxes by taking r objects from n objects is the number ofpermutations of n things taken r at a time.

Box nos. 1 2 3 ..... (r–1) r

No. of ways n n – 1 n – 2 n–(r–2) n–(r–1)

First place can be filled in n different ways

Second place can be filled in n – (2 – 1) = (n –1) ways

Third place can be filled in n – (3 – 1) = (n – 2) ways

The fourth place in n – (4 – 1) = (n – 3) ways

Similarly the rth place in n – (r –1) = (n – r + 1) ways

By Fundamental Principle r places can be

filled in n (n – 1) (n – 2) (n – 3) .... (n – r + 1) ways

This is the number of permutations of n things taken r at a time.

∴ nPr

= n (n – 1) (n – 2)......... (n – r + 1)

63

3. Factorial notationnP

r = n(n – 1)(n – 2) (n – 3) ........ (n – r + 1)

when r = nnP

n= n(n – 1) (n – 2) (n – 3)...... 3 x 2 x 1

Here R.H S is the Product of first n natural numbers and it is denoted by n!

∴ nPn = n!

Remember :

The product of first n consecutive natural numbers is denoted by n!or n and it is read as factorial n.

Similarly 5P5 = 5 x 4 x 3 x 2 x 1 4P

4= 4 x 3 x 2 x 1

= 5! = 120 = 4! = 245P5 = 5! 4P4 = 4!

nPr = n (n – 1) (n – 2) (n – 3) .................. (n – r + 1)

Multiply and divide by (n – r) (n – r – 1) ........... 3 x 2 x 1

123).....1rn)(rn(

]123).....rn)[(1rn).....(3n)(2n)(1n(nPr

n

××−−−××−+−−−−=

123).....1rn)(rn(

123).....2n)(1n(nPr

n

××−−−××−−=

∴ )!rn(!n

Prn

−=

Note (1)nP

r = n(n – 1)(n – 2)(n – 3) .... 3 x 2 x 1

When r = 1, 2, 3, 4............nP

1 = n

nP2

= n(n – 1)nP

3 = n(n – 1) (n – 2)

nP4 = n(n – 1) (n – 2) (n – 3)

Think :

n)!1n(

!n =−

64

Note (2)

)!rn(

!nPr

n

−=

When r = n

)!nn(

!nPn

n

−=

!0

!nPn

n =

1!n

!n

P

!n!0

nn ===

Worked Examples :1) Write the meaning of 6P

2

Solution : The number of arrangements of 6 different things taken 2 at a time is 6P

2

2) Find the value of 5P3

Solution : nPr

= n(n – 1) (n – 2) (n – 3) ..... (n – r + 1)

when n = 5, r = 35P

3= 5 (5 – 1) (5 – 2)

= 5 x 4 x 3

= 60

3) Evaluate 4P3 using formula

Solution: )!rn(

!nPr

n

−=

)!34(

!4P3

4

−=

!1

1234 ×××= 1

1234 ×××= = 24

4) In how many ways can 5 students be seated on a bench?

Solution : 5 students are arranged in a row.∴ Total number of arrangement = 5P

5

nPn = 5P

5

!5= 12345 ××××= 120= ways

Note (3)

)!rn(

!nPr

n

−=

When 0r =

)!0n(

!nP0

n

−=

!n

!n= = 1

∴ nP0 = 1

RecallnPn = nPn–1

RecallnP3 = n(n – 1) (n – 2)

Note:

1! = 1

65

5) Find the number of permutations of the letters of the word “MOBILE”.

Solution: The word MOBILE consists of 6 different letters

∴ Number of permutations = 6P6

n = 6nP

n= 6P

6

= 6!

= 6 x 5 x 4 x 3 x 2 x 1 = 720

6) How many 3 - digit numbers can be formed using the digits 2, 3, 4, 5, 6, 7, 8without repetitions?

Solution : There are 6 digits (∴ n = 6, r = 3)

∴ Number of 3 digit numbersnP

r= 6P

3

= 6 x 5 x 4 = 120

7) If nPn = 5040 find n

Solution : nPn

= 5040

n ! = 5040

n ! = 1 x 2 x 3 x 4 x 5 x 6 x 7

n! = 7!

∴ n = 7

8) If 11Pr = 990 find r

Solution : 11Pr = 990

11Pr = 11 x 10 x 9

11Pr = 11P

3

∴ r = 3

9) If nP2 = 90 find n

Solution : nP2 = 90

90)1n(n =−∴

910)1n(n ×=− n(n – 1) = 10(10 – 1)

∴ n = 10

1 50402 50403 25204 8405 2106 42 7

11 990

10 90

9

990 = 11 x 10 x 9

[∴ nP2 = n(n–1)]

66

10) If 5 nP3 = 4 n+1P

3 find n

Solution : nP3 = n(n – 1) (n – 2)

n+1P3

= (n + 1) n (n – 1) [Replacing n by (n + 1)]

∴ 5 )1n)(n)(1n(4)2n)(1n(n −+=−− Divide both sides by n (n-1)

)1n(4)2n(5 +=− 4n410n5 +=− 104n4n5 +=− ∴ n = 14

11) Find the number of arrangements of the letters of the word ARUN. How manyof these begin with A?

Solution: (i) Word ARUN has 4 different letters

∴ Number of arrangements = 4P4

= 4 !

= 4 x 3 x 2 x 1

= 24

(ii) For the words to begin with ‘A’ we fix the letter ‘A’ in the first place

A <–––––– 3P

3 ––––––>

After fixing the letter A in the first place, 3 letters R, U and N are left out.

In the remaining 3 places we arrange these 3 letters in 3P3 ways

∴ Number of arrangements = 3P3

= 3P3

= 3 x 2 x 1

= 6

12) How many 3-digit numbers can be formed by using the digits 3, 4, 5 and 6 withoutrepetitions? How many of these are even?

Solution : (i) There are 4 digits 3, 4, 5 and 6

∴ Number of 3-digit numbers = 4P3

= 4 x 3 x 2

= 24

67

(ii) Consider 3-blank boxes

Step (i): First fill up the unit’s place. This placecan be filled by any one of the digits4 or 6 in 2 ways.

Step (ii) : After filling this place, three digits are left out. So the Tens place can befilled in 3 ways.

Step (iii) : Similarly the hundreds place can be filled in the remaining 2 digits in 2 ways.

By Fundamental Principle, all the 3-places can be filled in 2 x 3 x 2 = 12 ways.

Exercise : 2.1

(1) Evaluate (a) 5! (b) 2! + 1! + 0! (c) 5P3

(2) Evaluate using formula : (a) 4P3

(b) 3P2

(c) 5P4

(3) Find 5P2

+ 4P0

– 3P1

(4) In how many ways 4 people out of 6 people can be seated out in a row for aphotograph?

(5) If 5Pr = 5 ! find r

(6) Find n if : (a) n! = 120 (b) n! = 720

(7) Find r if : (a) 13Pr = 156 (b) 8P

r = 336

(8) Find n if : (a) nP2 = 30 (b) nP

3 = 210

(9) Find the number of permutations of the letters of the words

(a) MILK (b) WORLD

(10) If nP4 = 12 nP

2 find n.

(11) How many 3 digit numbers can be formed using the digits 2, 3, 4, 5 and 6 withoutrepetitions? How many of these are even numbers?

(12) How many ways the letters of the word CHEMISTRY be arranged? How manyof these begin with M?

(13) How many 3-digit numbers are formed using the digits 0, 1, 2, 3 without repetitions?

(14) In how many ways can 7 different books be arranged in a shelf? In how manyways can we arrange three particular books so that they are always together?

H T U

2 3 2ways ways ways

68

A

D

B

C

COMBINATIONS

Consider an exampleThere are three boys ASHRITH (A), BHARATH (B) and CHETAN (C).

From these boys how many groups of two boys can be selected?

The selections are

1 A B

2 A C

3 B C

Three groups of selection can be made.

These selections are combinations

Example 2 :

There are 3 different books physics (P), Chemistry (C) and Mathematics (M)

In how many ways you can select them taking all at a time?

P C M

There is only one selection.

There is only one combination out of threebooks taken 3 at a time and is the only one wayof selection.

Example 3 :

There are four non-collinear points A, B, C, D on a plane. How many straight linescan be drawn by joining these points in pairs?

There are six straight lines.

AB, AC, AD, BC, BD, DC

These are the 6 ways of selection.

The number of combinations of 4 points taken 2 at a time is 6

Combination is mere a selection of different objects.

Observe :AB and BAAC and CA are not different selections.BC and CB

Observe :

PCM, PMC, CPM,CMP, MPC, MCP areall same selection

69

Example 4 :

How many committee of 3 members can be formed from a group of 4 people?

Let A, B, C, D be 4 people. Committee formed are

1 A B C

2 A C D

3 A D B

4 B C D

There are four committees. These are 4 combinations. The number of combinationsof 4 people taken 3 at a time is denoted by 4C

3

In general the number of combinations of ‘n’ things taken ‘r’ at atime is denoted by nCr.

Observe:

Note that : 3P2 = 3C

2 x 2!

1. To find the formula for number of combinations of ‘n’ things taken ‘r’ at atime.

Permutation of n things Selection of r Arrangement oftaken r at a time. = things out of n x r things.

∴ nPr = nC

r x r!

∴ !r

PC r

n

rn = ∴ !r)!rn(

!nCr

n

−=

Note:

ABC, ACB, BCA etc are one and the samecommittees

Note: 4C3 = 4

A, B, C BC2

AC

AB

BA

AB

BC

CB

AC

CA

(n)(r)

3C2 = 3

3P2 = 6

Given 3 things taken at atime

CombinationPermutation

Recall

)!rn(!n

Prn

−=

70

Note 1 :!r

PC r

n

rn =

When r = 2!2

PC 2

n

2n =

!2

)1n(nC2

n −=

Similarly!3

)2n)(1n(nC3

n −−=

!4

)3n)(2n)(1n(nC4

n −−−=

Note 2 : !r)!rn(

!nCr

n

−=

When r = n

!n)!rn(

!nCn

n

−=

!xn!0

!n= !xn1

!n= !n

!n= = 1

1Cnn =∴

Note 3 : !r)!rn(

!nCr

n

−=

When r = 0,

!0)!0n(

!nCo

n

−=

1!n

!n

×= = 1 1C0

n =∴

RecallnP2 = n (n – 1)

0! = 1

71

Note 4 : Verify �Cr < nP

r

Let n = 5, r = 3nP

r= 5P

3

5P3

= 5 x 4 x 3 = 60 ................. (1)nC

r= 5C

3

5C3

!3

P5 3= 123

345

××××= = 10 .............. (2)

From (1) and (2) : nCr < nP

r

Note 5 : nCr

= nCn–r

Let n = 5, r = 3nC

r= 5C

3

5C3

= !3

P35

123

345

××××=

6

60= ∴ 5C3

= 10 .............. (1)

nCn–r

= 5C5–3

= 5C2

5C2

= !2

P25

5C2

= 12

45

××

2

20= ∴ 5C3

= 10 ................. (2)

From (1) and (2): nCr = nCn–r

Note 6 : nC1 = n

nP1 = n

∴ nC1 = nP

1

In nPr and nCr, both n and r are positive integers and rn ≥ .

Remember :nCr ≤ nPr

Think :Why 4C5 is meaningless?

72

Worked Examples :

1) Evaluate 5C2 using formula

!r)!rn(

!nCr

n

−=

!2)!25(

!5C2

5

−=∴

!2x!3

!5= 12123

12345

×××/×/×/×/××=

2

4x5 2

//= = 10

2) Evaluate 100C99

using formula

Solution : 100C99

= 100C100-99

(∴ nCr = nC

n–r)

= 100C1

(∴ nC1 = n)

= 1

3) Find the number of combinations of the letters of the word “CAKE”.

Solution: The word CAKE has 4 different letters. Select all 4 letters at a time.

∴ The number of combinations = 4C4

= 1 ( ∴ nCn = 1)

4) There are 8 non-collinear points. How many straight lines can be drawn by joiningthese points?

Solution : A straight line is got by joining any two points on a plane. Thereare 8 non - collinear points.

∴ The number of straight lines

thus formed = 8C2

= !2

)18(8 −

!2

78×=

2

56=

= 28

!2

)1n(nC2

n −=

73

5) There are 7 non-collinear points. How many triangles can be drawn by joining thesepoints?

Solution: A triangle is formed by joining any three non-collinear points in pairs.

There are 7 non-collinear points

∴ The number of triangles formed = 7C3

!3

)27)(17(7 −−=

123

567

××××= = 35

6) If nC2 = 10 find n

Solution: nC2 = 10

10!2

)1n(n =−

n(n – 1) = 10 x 2! = 10 x 2 = 20

n(n – 1) = 5 x 4 = 5(5 – 1)

∴ n = 5 (By inspection)

7) If nC8 = nC

12 find n

Solution: nC8 = nC

12 nC

8 = nC

n–12(∴ nC

r = nC

n–r)

∴ 8 = n – 12

8 + 12 = n ∴ n = 20

8) If 6Pr = 360 and 6C

r = 15 find r

Solution: nPr = nC

r x r !

6Pr = 15 x r!

360 = 15 x r!

∴ 2415

360!r ==

= 4 x 3 x 2 x 1

r! = 4! ∴ r = 4

12!2 ×=

74

9) A Box contains 5 Blue and 4 Red marbles. In how many ways can 4 marblesbe drawn so as to include 2 red marbles?

Solution: There are 4 red marbles. The number of ways in which 2 red marblescan be drawn from 4 red marbles is 4C

2 ways

∴ 4C2 !2

34×= 12

34

××=

2

12= = 6

Next select 2 marbles from 5 blue marbles in 5C2

ways

∴ 5C2

!2

45×= 12

45

××=

2

20= 10=

∴ The total number of ways = 6 x 10 = 60

10) How many diagonals can be drawn in a pentagon?

Solution. A pentagon has 5 sides. We obtain thediagonals by joining the vertices in pairs.

∴ Total number of sides and diagonals

= 5C2

!2

45×= = 5 x 2 = 10

This includes its 5 sides also.

∴ Diagonals = 10 – 5 = 5

Hence the number of diagonals = 10 – 5 = 5

Remember :If a polygon has n sides, then it has (nC

2– n) diagonals.

11) In how many ways can 4 people be selected from a group of 6 among whichAshok is one? How many of these include Ashok?

Solution: (i) There are 6 people.

∴ Number of 4 people groups = 6C4

= 6C6–4

= 6C2

!2

56×= 2

30= 15=

Activity

Draw a pentagon and itsdiagonals

Diagonals are AC, AD,BE, DB, CE∴ 5 diagonals

E

A B

D

C

75

(ii) Ashok is always included.

Now form groups of 3 people out of 5 people. This can be done in 5C3

ways.∴ 5C

3= 5C

5–3 ways

= 5C2

!2

45×=

2

20

12

45 =××= = 10

12) From 8 gentlemen and 4 ladies, a committee of 6 is to be formed. In how manyways can this be done so that the committee contains exactly 2 ladies?

Condition: Exactly two ladies.

Ladies Gentlemen Number of ways 4 8

2 4 4C2 x 8C

44C

2 = 6

= 6 x 70 8C4 = 70

= 420

13) There are 6 bowlers and 9 batsmen in a cricket club. In how many ways cana team of 11 be selected so that the team contains at least 4 bowlers?

Solution:

Possibilities Bowlers Batsmen Number of ways

6 9

(1) 4 7 6C4

x 9C7

(2) 5 6 6C5

x 9C6

(3) 6 5 6C6

x 9C5

6C4

x 9C7

= 15 x 36 = 5406C

5x 9C

6= 6 x 84 = 504

6C6

x 9C5

= 1 x 126 = 126

Total = 1170

76

Exercise : 2.2

1) Evaluate : (a) 5C4

(b) 100C98

2) Evaluate using formula : (a) 5C3

(b) 7C2

3) Find n if nC8 = nC

5

4) Find n given that (n + 1)C3

= 4 x nC2

5) If nPr = 336 and nC

r = 56 find n and r

6) A school has 8 teachers. One of them is the Headmaster (a) How many committeesof 5 can be formed? (b) How many of them have the Headmaster as a member?

7) There are 7 girls in a class room. Ashwini is one of them. In how many waysa committee of 5 be formed so as to exclude Ashwini?

8) There are 8 members in a club of which A and B are two members. A Committeeof 5 is to be formed. How many of these will include A and exclude B?

9) There are 5 bowlers and 10 batsmen in a cricket club. In how many ways cana team of 11 be selected so that the team contains exactly 3 bowlers?

10) From 8 gentlemen and 5 ladies a committee of 6 is to be formed. In how manyways can this be done so that the committee contains at least 3 ladies?

11) There are 20 non-collinear points in a plane. How many (a) straight lines (b) Trianglescan be drawn by joining these points?

JUST FOR FUN‘Tower of Brahma’

‘Tower of Brahma’ is a temple in the Indian cityof Benares. This tower, the description reads,consists of 64 disks of gold, now in the processof being transferred by the temple priests. Beforethey complete their task, it was said, the templewill crumble into dust and the world will vanishin a clap of thunder. The disappearance of theworld may be questioned, but there is littledoubt about the crumbling of the temple. Theformula 264 – 1, yields the 20-digit number18,446,744,073,709,551,615. Assuming that thepriests worked night and day, moving one diskevery second, it would take them many thousandsof millions of years to finish the job.

If we consider a tower of three disks only, thenthe number of moves will be 23 – 1 = 7. So itcan be solved by moving the disks in thefollowing order : ABACABA

If we consider a tower of four disks then thenumber of moves will be 24 – 1 = 15 i.e,ABACABADABACABA.