# Permutation and Combination Aptitude Questions and Answer – Quantitative Aptitude

Permutation and Combination Aptitude Questions and Answer – Tips & Tricks

Factorial:

Factorial can defined, the product of the number by its successor till it reaches to one.

n! = n (n – 1) (n – 2) ….. 1

Permutation:

Permutation, the number of ways particular set can be arranged, when order of the arrangement matters. A combination lock can be called a permutation lock. for example

• We have three letters a, b and c, We have to arrange two letters at a time. So, in this example, the permutations of two letters = ab, ba, bc, cb, ac, and ca.
• If we have to arrange all letters (a,b,c), the permutation would be abc, acb, bac, bca, cab, and cba.

nPr = n(n – 1)(n – 2)(n – 3) … (n – r + 1) = (n! / (n – r)!)

Combinations:

The number of ways a particular set can be arranged, where order of the arrangement does not matter which means for a combination of the n number of things there may be different orders. For Example

• Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.
• 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.

nCr = (n! / ((r!) * (n – r)!)

Note:

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

Permutation and Combination Aptitude Questions and Answer With Detailed ExplanationQuantitative Aptitude

## Permutation and Combination

 Question 1
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 5 B 10 C 15 D 20
Question 1 Explanation:
Explanation:
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 ten place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Therefore, Required Number of Numbers = (1 * 5 * 4) = 20
 Question 2
Using all the letters of the word "THURSDAY", how many different words can be formed?
 A 8 B 8! C 7 D 7!
Question 2 Explanation:
Explanation:
Total Number of letters = 8
Using these letters the number of 8 letters words formed is 8P8 = 8!
 Question 3
Six points ar marked on a straight line and five points are marked on another line, Which is parallel to the first lie. How many straight lines, including the first two, can be formed with these points?
 A 30 B 32 C 34 D 36
Question 3 Explanation:
Explanation:
We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear except those that can be selected out of these 6 points are collinear.
Hence, the required number of stright lines
=11C2 - 6C2 - 52 + 1 + 1
= 55 - 15 - 10 + 2
Ans = 32
 Question 4
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 in the committee. In how many ways can it be.
 A 702 B 624 C 756 D 812
Question 4 Explanation:
Explanation:
From a group of 7 men and 6 women, 5 persons are to be selected with at least 3 men.
Hence, we have the following 3 options

We can select 5 men ...(option 1)
Number of ways to do this = 7C5

We can select 4 en and 1 women.....(option 2)
Number of ways to do this = 7C4 * 6C1

We can select 3 men and 2 women.....(option 3)
Number of ways to do this = 7C3 * 6C2

Total number of ways
7C5 + (7C4 * 6C1) + (7C3 * 6C2)
7C2 + (7C3 * 6C1) + (7C3 * 6C2)

nCr = nC(n-r)

= = 21 + 210 + 525
Ans = 756
 Question 5
A coin is tossed 3 times. Find out the number of possible outcomes.
 A 2 B 8 C 1 D None of these
Question 5 Explanation:
Explanation:
When a coin is tossed once, there are two possible outcomes: Head(H) and Tale(T)
Hence, When a coin is tossed 3 times, the number of possible outcomes
= 2 * 2 * 2 Ans = 8
 Question 6
There are 6 persons in an office. A group consisting of 3 persons has to be formed. In how many ways can the group be formed?
 A 10 B 20 C 30 D 40
Question 6 Explanation:
Explanation:
Number of ways in which the group can be formed Ans = 20
 Question 7
Find the value of 50P2
 A 4500 B 3260 C 2450 D 1470
Question 7 Explanation:
Explanation: Question 8
In how many ways can 10 engineers and 4 doctors be seated at a round table without any restriction?
 A 12! B 10! C 13! D 14!
Question 8 Explanation:
Explanation:
Number of circular permutations(arrangements) of n distinct things = (n-1)!
Here, n = 10 + 4 = 14
Hence, Number of arrangements possible
=(14 - 1)! = 13!
 Question 9
In how many ways can 11 persons be arranged in a row such that 3 particular persons should always be together?
 A 9! B 9! * 3! C 11! * 3! D 11!
Question 9 Explanation:
Explanation:
Given that 3 particular persons should always be together. Hence, just group these 3 persons together and consider as a single person.
Therefore, We can take total number of persons as 9. These 9 persons can be arranged in 9! ways.
We had grouped 3 persons together. These 3 persons can be arranged among themselves in 3! ways.
Hence, required number of ways
Ans=9! * 3!
 Question 10
Out of 7 Consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
 A 210 B 1050 C 25200 D 21400
Question 10 Explanation:
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
=(7C3 * 4C2) = 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!
=5 * 4 * 3 * 2 * 1
=120
Therefore, Required Number of ways = (210 * 120) = 25200
There are 10 questions to complete.
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