# Tips to solve Permutation and Combination Problems 1. 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:

1. We define 0! = 1.
2. 4! = (4 x 3 x 2 x 1) = 24.
3. 5! = (5 x 4 x 3 x 2 x 1) = 120.
2. Permutations:The different arrangements of a given number of things by taking some or all at a time, are called permutations.Examples:
1. All permutations (or arrangements) made with the letters abc by taking two at a time are (abbaaccabccb).
2. All permutations made with the letters abc taking all at a time are:
( abcacbbacbcacabcba)
3. Number of Permutations:Number of all permutations of n things, taken r at a time, is given by:
 nPr = n(n – 1)(n – 2) … (n – r + 1) = n! (n – r)!

Examples:

1. 6P2 = (6 x 5) = 30.
2. 7P3 = (7 x 6 x 5) = 210.
3. Cor. number of all permutations of n things, taken all at a time = n!.
4. An Important Result:If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,
such that (p1 + p2 + … pr) = n.

 Then, number of permutations of these n objects is = n! (p1!).(p2)!…..(pr!)
5. 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:
1. 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.
2. All the combinations formed by abc taking abbcca.
3. The only combination that can be formed of three letters abc taken all at a time is abc.
4. Various groups of 2 out of four persons A, B, C, D are:

AB, AC, AD, BC, BD, CD.

5. Note that ab ba are two different permutations but they represent the same combination.
6. Number of Combinations:The number of all combinations of n things, taken r at a time is:
 nCr = n! = n(n – 1)(n – 2) … to r factors . (r!)(n – r)! r!

Note:

1. nCn = 1 and nC0 = 1.
2. nCr = nC(n – r)

Examples:

 i.   11C4 = (11 x 10 x 9 x 8) = 330. (4 x 3 x 2 x 1)
 ii.   16C13 = 16C(16 – 13) = 16C3 = 16 x 15 x 14 = 16 x 15 x 14 = 560. 3! 3 x 2 x 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?
 A. 564 B. 645 C. 735 D. 756 E. None of these

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only). Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
 = 7 x 6 x 5 x 6 x 5 + (7C3 x 6C1) + (7C2) 3 x 2 x 1 2 x 1
 = 525 + 7 x 6 x 5 x 6 + 7 x 6 3 x 2 x 1 2 x 1
= (525 + 210 + 21)
In how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together?
 A. 360 B. 480 C. 720 D. 5040 E. None of these

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.

= 756.

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

Correct! Wrong!

#### Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Correct! Wrong!

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

Correct! Wrong!

#### In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

Correct! Wrong!

#### 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?

Correct! Wrong!

QUIZ ON PERMUTATIONS & COMBINATIONS
GREAT JOB!!!