# Permutations and Combinations III: Permutations and Combinations Questions for Bank Exams

In this post, we will solve few frequently asked Permutations and Combinations Questions to save time in Bank Exams

Permutations and Combinations Questions need you to solve problems that are based on arrangements. Such questions which appear in bank exams can be solved with the help of permutations and combinations formulas. In this post, we will discuss few frequently asked permutations and combinations questions that are based on- conditional permutations and combination. These questions will be solved using permutations and combinations formulas to save time in Bank Exams.
Condition based Permutation and Combinations Questions

Example 1: Permutations and Combinations Questions

Question: How many different words can be formed using the letters of the word ‘EDUCATION” such that word always starts with ‘D’?

Solution:
Step 1:
There are 9 letters in the word EDUCATION. Using the letters of EDUCATION, 9! words can be formed. But the questions states that all the words should start with D.
Hence, 8 positions keep on changing and 1 position is fixed.
1 x 8!

Step 2:
1 x 8! = 1 x (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8) = 40,320

Therefore, 40,320 words can be formed starting from D and using the letters from EDUCATION.

Example 2: Permutations and Combinations Questions

Question: How many different words that can be formed using the letters of the word EDUCATION such that the word always ends with a vowel?

Solution:
Step 1: There are 9 letters in the word EDUCATION. Using the letters of EDUCATION, 9! words can be formed. But the questions states that all the words that are formed should always end with a vowel.

Step 2:
There are 5 vowels in the word (E, U, A, I, O) = 5

Hence, 8 positions keep on changing as 1 position is fixed = 8!

Step 3:
5 x 8! = 5 x (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8) = 5 x 40,320 = 2,01,600

Therefore, 2,01,600 words can be formed from the letters of EDUCATION which starts with a vowel.

Example 3: Permutations and Combinations Questions

Question: How many different words can be formed using the letters of the EDUCATION such that the word always begins with the letter ‘A’ and ends with a consonant?

Solution:
Step 1: There are 9 letters in the word EDUCATION. Using the letters of EDUCATION, 9! words can be formed. But the questions states that all the words that are formed should always begin with A and end with a consonant.

Step 2:
One place is fixed as the words should begin with A: 1
There are 7 positions that keep changing: 7!
The last place is fixed, as the words should end with a consonant: 4

Step 3:
1 x 7! X 4 = 7! X 4 = (1 x 2 x 3 x 4 x 5 x 6 x 7) x 4 = 20160

Therefore, 20160 words that begin with A and ends with a consonant can be formed using the letters from EDUCATION.
Permutations and Combinations Questions when repetition is allowed and when repetition is not allowed

Example 4: Permutations and Combinations Questions

Question: How many 5 digits numbers can be formed with the digits 2, 4, 5, 8 and 9 when
(i) When the repetition is not allowed             (ii) When repetition is allowed

Solution:
(i) When the repetition is not allowed

Answer: When the repetition is not allowed there are 5 positions and there is no repetition of any number.
Therefore, 5! = 120, 5 digit numbers can be formed using the digits 2,4,5,8 and 9.

(ii)  When repetition is allowed

Answer: As repetition is allowed all the positions can be filled in 5 different ways.

Therefore, 55 = 3125, 5 digit numbers can be formed using the digits 2, 4, 5, 8 and 9.

Example 5: Permutations and Combinations Questions

Question: How many 4 digit numbers can be formed with the digits 0, 1, 3 and 6?

Solution:
Step 1:
There are four digits that are given in the question – 0,1,3,9
We are suppose to find out how many 4 digit numbers that can be formed using the digits.
If 0 is placed in the thousands place then the number will become a 3-digit number. So the first position can only be acquired by the 3 digits i.e. 1, 3 and 9.

Step 2:
Using the principle of multiplication (m x n ways) we get,
3 x 3 x 2 x 1 = 18

Therefore, 18 four-digit numbers can be formed using the digits 0, 1, 3 and 6.

Permutations and Combinations Questions for Practice!

Question: In how many different ways can the letters of the word DETAIL be arranged in such a way that the vowels occupy only the odd positions?
(1) 60         (2) 54          (3) 120       (4) 36          (5) None of these

Question: In how many different ways can the letters of the word ‘RUMOUR’ be arranged?
(1) 360        (2) 180         (3) 240         (4) 720             (5) None of these