## 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
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 [4] + [(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

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