Problems on Permutation and Combination
 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
1. 
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: Option D
Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

2. 
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
 
Answer: Option C
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.

3. 
In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
 
Answer: Option D
Explanation:
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
Required number of ways = (2520 x 20) = 50400.

4. 
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
 
Answer: Option C
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Required number of ways = (210 x 120) = 25200.

5. 
In how many ways can the letters of the word 'LEADER' be arranged?
 
Answer: Option C
Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

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