# Permutations and Combinations V: Important Permutations and Combinations Questions for Bank Exams

In this post, we will discuss few important Permutations and Combinations Questions that are asked in Bank Exams

Permutations and Combinations Questions are the questions that ask you to find different ways in which you can arrange or select objects. Permutations and Combinations Questions are complicated and often candidates find it difficult to solve them in Bank Exams. This is the fifth post on Permutations and Combinations. In this post, we will discuss few important basic, complex, and conditional combinations questions that will help you save time in Bank Exams.
Example 1: Permutations and Combination Questions

Question: In how many different ways can a committee of 8 persons be formed out of 5 men and 3 women?
Solution:
Step 1:
When we form a committee, we only need to select persons irrespective of their gender. Their arrangement or their order of selection is not required.
5 men + 3 women = 8 (n)

Step 2:
Step 3:
8c8 = 8! / (8 – 8)! x 8! = 1

Therefore, 1 committee of 8 persons can be formed out of 5 men and 3 women.
Hence,

Example 2: Permutations and Combination Questions

Question: In how many ways can a cricket team of 11 players be chosen out of total 14 players?

Solution:
Step 1:
Step 2:
n = 14!
r = 11!

Step 3:
14c11 = 14! / [(14 – 11)! x 11!]  = 14! / (3! x 11!)
14c11 = (14 x 13 x 12 x 11!) / 6 x 11! (11! In the numerator and denominator is cancelled and 12 goes 2 times in 6, hence 6 and 12 are also cancelled)
14c11 = 364 ways

Therefore, there are 364 ways in which a team of 11 players can be chosen from 14 players.
Example 3: Permutations and Combination Questions

Question: Out of 10 men, there are 4 doctors, 3 teachers and 3 lawyers and out of 8 women, there are 3 doctors, 3 dancers and 2 lawyers. In how many ways can a committee of 5 persons be formed such that there are 3 doctors and 2 lawyers?

Solution:
Step 1:
This question is a case of a conditional combination. We need to form a committee such that there are 3 doctors and 2 lawyers in them.

Step 2:
4 men doctors + 3 women doctors = 7 doctors
Since we need to select 3 doctors out 7 doctors, this can be done in 7c3 ways.

Step 3
3 Men lawyers + 2 women lawyers = 5 lawyers
Since we need to select 2 lawyers out 5 lawyers, this can be done in 5c2 ways.

Step 4:
To find the total number of ways in which a committee can be formed of 5 persons,
We need to multiply,
7c3 x 5c2

Step 5:
7c3 = 7! / (7 -3)! x 3! = 5040 /12 = 420
5 c2 = 5! / (5 -2)! x 2! = 120 /12 = 10

Step 6:
420 x 10 = 4200 ways

Therefore, there are 4200 ways from which a committee of 5 persons can be formed such that there are 3 doctors and 2 lawyers in it.

Example 3: Permutations and Combination Questions

Question: In how many ways can a cricket team of 11 players be chosen out of 8 batsmen and 6 bowlers such that
(i) there are 7 batsmen                                (ii)  there are 5 bowlers

Solution:
(i) there are 7 batsmen
Step 1:
There are 8 batsmen out of which we need to choose 7 batsmen = 8c7
Since we need to form a team of 11 players that indicates that there are 4 bowlers in the team = 6c4

Step 2:
8c7 = 8!/ (8 - 7)! x 7! = 8/1 = 8
6c4 = 6!/ (6 - 4)! x 4! = 6!/ (2! x 4!) = (5 x6)/ 2! = 15

Step 3:
8 x 15 = 120

Therefore, there are 120 ways in which a team of 11 players can be formed such that there are 7 batsmen in the team.

(ii)  there are 5 bowlers
Step 1:
There are 6 bowlers out of which we need to choose 5 bowlers = 6c5
Since we need to form a team of 11 players that indicates that there are 6 batsmen in the team = 8c6

Step 2:
6c5 = 6!/ (6 - 5)! x 5! = 6!/ 5! = 6
8c6 = 8!/ (8 - 6)! x 6! = 8!/ (2! x 6!) = (7 x 8)/ 2= 28

Step 3:
6 x 28 = 168

Therefore, there are 168 ways in which a team of 11 players can be formed such that there are 5 bowlers in the team.
Example 4: Permutations and Combination Questions

Question: In how many ways can a person choose one or more books out of 5 different subject books?

Solution:
Conventional Method:
Step 1:
The word “choose” indicates that this is a combination question.
There are 5 books and all these 5 books can be selected in 5 different ways.
Hence, 5c1 + 5c2 + 5c3 + 5c4 + 5c5
= 5 + 10 + 10 + 5 + 1 = 31 ways.

Therefore, there are 31 ways in which a student can select one or five books out of 5 different subject books.

Smart Method:
Step 1:
There are 5 different books and a student can select a book or reject a book.
Hence, there are 2 possibilities with each book. That means that there are 25 different ways in which a person can choose one or more books.
But there is one way in which a person can reject all the books at a time.
25 - 1 = 32 – 1 = 31 ways

Therefore, there are 31 ways in which a student can select one or five books out of 5 different subject books.

Permutations and Combinations Questions for Practice

Question:  A dance group is such that the number of ways of selecting 5 persons is same as the number of ways of selecting 7 persons. In how many ways can 9 persons be selected from this group?
1) 180               2) 220            3) 200              4) 190              5) None of these

Question: In a box of 4 pens and 6 rulers, 4 items are to be selected. In how many different ways it can be selected such that at least one pen should be there?
1) 195               2) 120            3) 240              4) 96                5) None of these