## Aptitude Permutation And CombinationPage 1

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:
We define 0! = 1.
3! = (3 x 2 x 1) = 6.

2. Permutations:
The different arrangements of a given number of things by taking some or all at a time, are called permutations.
Examples:
All permutations made with the letters a, b, c taking all at a time are:
( abc, acb, bac, bca, cab, cba)

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:
66P2 = (6 x 5) = 30.
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:
i. 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.
ii. All the combinations formed by a, b, c taking ab, bc, ca.
iii. The only combination that can be formed of three letters a, b, c taken all at a time is abc.
iv. Various groups of 2 out of four persons A, B, C, D are:
AB, AC, AD, BC, BD, CD.
v. 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!/(r!)(n - r)! = n(n - 1)(n - 2) ... to r factors/r! .

Note:
nCn = 1 and nC0 = 1.
nCr = nC(n - r)
Examples:
i. 11C4 = (11 x 10 x 9 x 8)/(4 x 3 x 2 x 1) = 330.

ii. 16C13 = 16C(16 - 13) = 16C3 = 16 x 15 x 14/3! = 16 x 15 x 14/3 x 2 x 1 = 560.

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)256 (b)520 (c)756 (d)840
Answer is: CWe 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).
Required number of ways = [(7 x 6 x 5 )/(3 x 2 x 1)] x [(6 x 5)/(2 x 1)] + (7C3 x 6C1) + (7C2).
Required number of ways = 525 + [(7 x 6 x 5)/(3 x 2 x 1)] x 6 + [(7 x 6)/(2 x 1)].
Required number of ways = 525 + 210 + 21 = 756

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? (a)480 (b)512 (c)720 (d)840
Answer is: CThe 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? (a)4500 (b)50400 (c)65010 (d)72000
Answer is: BIn 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.
Number of ways arranging these letters = 7!/2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5!/3! = 20 ways.
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? (a)24500 (b)25200 (c)27300 (d)28100
Answer is: BNumber of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3 x 4C2) = [(7 x 6 x 5)/(3 x 2 x 1)] x [(4 x 3)/(2 x 1) = 210
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves = 5!
Number of ways of arranging 5 letters among themselves = 5 x 4 x 3 x 2 x 1
Number of ways of arranging 5 letters among themselves = 120.
Required number of ways = (210 x 120) = 25200.

5.

 In how many ways can the letters of the word LEADER be arranged? (a)120 (b)260 (c)320 (d)360
Answer is: DThe word LEADER contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Required number of ways = 6!/((1!)(2!)(1!)(1!)(1!)) = 360.

6.

 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? (a)200 (b)209 (c)305 (d)309
Answer is: BWe may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
Required number of ways = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
Required number of ways = (6 x 4) + [(6 x 5)/(2 x 1)] x [(4 x 3)/(2 x 1)] + [(6 x 5 x 4)/(3 x 2 x 1)] x  + [(6 x 5)/(2 x 1)]
Required number of ways = (24 + 90 + 80 + 15)
Required number of ways = 209

7.

 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? (a)15 (b)20 (c)25 (d)30
Answer is: BSince each desired number is divisible by 5, so we must have 5 at the unit place. So, there is one 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.

8.

 How many words can be formed by using all letters of the word BIHAR ? (a)120 (b)60 (c)180 (d)260
Answer is: AThe word BIHAR contains 5 different letters.
Required number of words = 5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120