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**Permutations and Combinations**

- Factorial Notation:Let
*n*be a positive integer. Then, factorial*n*, denoted*n*! is defined as:n! = n(n - 1)(n - 2) ... 3.2.1.Examples:- We define 0! = 1.
- 4! = (4 x 3 x 2 x 1) = 24.
- 5! = (5 x 4 x 3 x 2 x 1) = 120.

- Permutations:The different arrangements of a given number of things by taking some or all at a time, are called permutations.Examples:
- All permutations (or arrangements) made with the letters
*a*,*b*,*c*by taking two at a time are (*ab*,*ba*,*ac*,*ca*,*bc*,*cb*). - All permutations made with the letters
*a*,*b*,*c*taking all at a time are:

(*abc*,*acb*,*bac*,*bca*,*cab*,*cba*)

- Number of Permutations:Number of all permutations of
*n*things, taken*r*at a time, is given by:^{n}P_{r}=*n*(*n*- 1)(*n*- 2) ... (*n*-*r*+ 1) =*n*!( *n*-*r*)!Examples:^{6}P_{2}= (6 x 5) = 30.^{7}P_{3}= (7 x 6 x 5) = 210.- Cor. number of all permutations of
*n*things, taken all at a time =*n*!.

- An Important Result:If there are
*n*subjects of which*p*_{1}are alike of one kind;*p*_{2}are alike of another kind;*p*_{3}are alike of third kind and so on and*p*_{r}are alike of*r*^{th}kind,

such that (*p*_{1}+*p*_{2}+ ...*p*_{r}) =*n*.Then, number of permutations of these *n*objects is =*n*!( *p*_{1}!).(*p*_{2})!.....(*p*_{r}!) - Combinations:Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.Examples:
- Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.Note: AB and BA represent the same selection.
- All the combinations formed by
*a*,*b*,*c*taking*ab*,*bc*,*ca*. - The only combination that can be formed of three letters
*a*,*b*,*c*taken all at a time is*abc*. - Various groups of 2 out of four persons A, B, C, D are:AB, AC, AD, BC, BD, CD.
- Note that
*ab**ba*are two different permutations but they represent the same combination.

- Number of Combinations:The number of all combinations of
*n*things, taken*r*at a time is:^{n}C_{r}=*n*!= *n*(*n*- 1)(*n*- 2) ... to*r*factors. ( *r*!)(*n*-*r*)!*r*!Note:^{n}C_{n}= 1 and^{n}C_{0}= 1.^{n}C_{r}=^{n}C_{(n - r)}

Examples:i. ^{11}C_{4}=(11 x 10 x 9 x 8) = 330. (4 x 3 x 2 x 1) ii. ^{16}C_{13}=^{16}C_{(16 - 13)}=^{16}C_{3}=16 x 15 x 14 = 16 x 15 x 14 = 560. 3! 3 x 2 x 1

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**Example:**

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

Answer:

Required number of ways= (

= 11760.

^{8}C_{5}x^{10}C_{6})= (^{8}C_{3}x^{10}C_{4})= | 8 x 7 x 6 | x | 10 x 9 x 8 x 7 | |

3 x 2 x 1 | 4 x 3 x 2 x 1 |

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**Example:**

In how many ways can the letters of the word 'LEADER' be arranged?

Answer:

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

Required number of ways = | 6! | = 360. |

(1!)(2!)(1!)(1!)(1!) |

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**Example:**

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

Answer:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways | = (^{7}C_{3} x ^{6}C_{2}) + (^{7}C_{4} x ^{6}C_{1}) + (^{7}C_{5}) | |||||||||||

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= (525 + 210 + 21) | ||||||||||||

= 756. |

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**Example:**

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Answer:

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

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**Example:**

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

Answer:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters = | 8! | = 10080. |

(2!)(2!) |

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters = | 4! | = 12. |

2! |

Required number of words = (10080 x 12) = 120960.

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