# Permutation and Combination Aptitude Questions

Hey, Coders! In this post, we will discuss one of the most confusing aptitude topics i.e Permutations and Combinations. We will also discuss Permutation and Combination aptitude questions in this post.

Everyone finds this topic difficult at first, but it becomes easy as one proceeds and keeps practicing.

The other day, I wanted to travel from Mumbai to Delhi by train. There is no
direct train from Mumbai to Delhi, but there are trains from Mumbai to Ratlam
and from Ratlam to Delhi.

From the railway timetable, I found that there are two trains
from Mumbai to Ratlam and three trains from Ratlam to Delhi. Now, in how many
ways can I travel from Mumbai to Delhi?
These are certain counting problems that come under the branch of Mathematics called combinatorics or combinations.

Now, we will proceed to understand the permutations and combinations principle.

## Overview

Remember the 3 golden rules when it comes to P&C-

1. When the order doesn’t matter, it is a Combination.
2. When the order does matter it is a Permutation!
3. Never forget Rule 1 & 2

Let’s make it easy to understand through an example.

“My pizza is a combination of corns, olives, cheese, and capsicum, etc.” Here, we don’t care what order the things are in, they could also be “cheese, olives, corns” or “corns, cheese, olives”, its the same pizza!

“The combination to the safe is 678”. Now we do care about the order. “876” won’t work, nor will “687”. It has to be exactly 6-7-8.

Solving a permutation or combination problem is a two step process,

1. Recognizing the problem type i.e permutation or combination.

2. Using formulas or models to count the possibilities.

The problem that students face is how to identify whether the problem is for permutation or for a combination?

The best way to do is switch the items of the question and checking is the result the same?

If you need to pick a Principal and Head-Teacher, then switching them changes the result. So, picking a Principal and Head-Teacher from a set of candidates is permutation.

If you want to choose 2 players for a fight, then switching the order doesn’t change the result: either way, they are fighting each other. Hence, it’s a combination.

According to the given problem, one should check if the problem allows repitition or not.

## What is Combination?

The combination is basically a process to select things where order does not matter. Here we need to arrange the digits, numbers, and alphabets taking some or all at a time. We can represent it as nCr.

This is how the lottery work. We can draw the numbers one at a time, and if we have the lucky numbers (no matter what order) we win!

The easiest way to explain it is to:

1. Assume that the order does matter (ie permutations),
2. Then alter it so the order does not matter.

## What is Permutation?

Permutation is basically a process where order matters. Here we need to arrange the digits , numbers , and alphabets taking some or all at a time. We represent it as nPr . There are basically two types of permutations:

1. Repetition: When we can repeat a certain number of things.
2. No Repetition: For example the first 3 winners in a painting competition. Here, one can’t be first and second.

### 1. Permutations with Repetition

These are the easiest to calculate. When a thing has n different types i.e we have n choices each time!

For example: choosing 3 of those things, the permutations are:

n × n × n(n multiplied 3 times)

More generally: choosing r of something that has n different types, the permutations are:

n × n × … (r times)

(In other words, there are n possibilities for the first choice, THEN there are n possibilities for the second choice, and so on, multiplying each time.)

Thus, we can say that – n × n × … (r times) = nr

### 2. Permutations without Repetition

In this case, we have to reduce the number of available choices each time. Let’s take an example. In how many ways can one arrange 15 things if we don’t repeat the things?

After choosing, say, number “14” we cannot choose it again.

So, our first choice has 15 possibilities, and our next choice has 14 possibilities, then 13, 12, 11, … etc. And the total permutations are:

15 × 14 × 13 × … = 1,307,674,368,000

But maybe we don’t want to choose them all, just 2 of them, and that is then:

15 × 14 = 210

## Permutation and Combination aptitude questions

### Solved Examples – Combinations

Q1.Suppose 7 students stay in a hall in a hostel and they are given 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?

Solution: Let the beds be numbered 1 to 7.
Case 1: Suppose Anju is allotted bed number 1. Then, Parvin cannot be allotted bed number 2.
So Parvin can be allotted a bed in 5 ways. After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways. So, in this case, the beds can be allotted in 5 x 5! =600ways.

Case 2: Anju is allotted one of the beds of number 2,3,4,5 or 6. Parvin cannot be allotted the beds on the right-hand side and left-hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin. Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways. Therefore, in each of these cases, the beds can be allotted in 4× 5! = 480 ways.
∴ The beds can be allotted in
(2×600+5×480)ways= (1200+2400)ways= 3600ways

Q2. In how many ways can an animal trainer arrange 5 lions and 4 tigers in a row so that no two lions are together?

Solution: We can arrange them in the following way :
L T L T L T L T L
We can arrange the 5 lions in the 5 places marked ‘L’.
We can do this in 5! ways.
We can arrange the 4 tigers in the 4 places ‘T’.
We can do this in 4! ways.
Therefore, we can ararnge the lions and the tigers in 5!x4! =2880 ways.

Q3. 12 points lie on a circle. How many cyclic quadrilaterals we can draw by
using these points?

Solution: For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4
points out of 12 points is 12`C4` = 495. Therefore, we can draw 495 quadrilaterals.

### Solved Examples – Permutations

Q1. How many arrangements of the letters of the word ‘BENGALI’ can be made
(i) if the vowels are never together.
(ii) if the vowels are to occupy only odd places.

Solution: There are 7 letters in the word ‘Bengali; of these 3 are vowels and 4 consonants.
(i) Considering vowels a, e, i as one letter, we can arrange 4+1 letters in 5! ways in each
of which vowels are together. We can arrange these 3 vowels among themselves in 3!
ways.
∴ A total number of words = 5! × 3!
= 120 × 6 = 720
(ii) There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in 3
4`P3` ways
We can arrange 4 constants in
4`P`4 ways.

∴ Number of words =24 x 24 = 576.

Q2.How many three-digit numbers can we form by using the digits in 735621, if repetition is not allowed?

Solution: nPr = n! / (n-r)!

6P=  6! / (6-3)!

6 P= 6!/3!

6P= 120

Q3. Find the number of different words that can be formed from the word ‘SUCCESS’.

Solution : No. of Permutation = n! / p! × q!, where p = of one type , q = ( of another type ).
No. of Permutation = 7!/ 3! × 2!

No. of Permutation = 420

### Problems for practice

Q1.In how ways can we select 6 persons from 4 grade 1 and 7 grade II officers, so as to
include at least two officers from each category?

Q2.We need to form a committee of 5 out of 6 boys and 4 girls. In how many ways can we form if we take :
(a) 2 girls.
(b) at least 2 girls.

Q3.The English alphabet has 5 vowels and 21 consonants. What is the maximum number of
words, that one can form from the alphabet with 2 different vowels and 2 different
consonants?

Q4.From 5 consonants and 5 vowels, how many words can we form using 3 consonants
and 2 vowels?

Q5. In a school annual day function, they decided to plan a program. According to the plan,
there would be 3 short plays, 6 recitals, and 4 dance programs. However, the chief
guest invited for the function took a much longer time than expected to finish his speech.

To finish in time, it was a plan that only 2 short plays, 4 recitals, and 3 dance programs
would be performed. How many choices were available to them?

(a) if we perform the programs in any order?
(b) if we perform programs of the same kind at a stretch?
(c) if we perform programs of the same kind at a stretch and considering the
order of performance of the programs of the same kind?

Q6.In how many ways can arrange the letters of the word ‘ACCOMPLISHMENT’ such that

1. All vowels are together.
2. Vowels are never together.

Q7. We have to form a committee of 5 persons from 6 men and 4 women. In how many ways can we do this?

1. We include at least 2 women?
2. We include at most 2 women?

## Conclusion

OMG! that was a lot to absorb, we would recommend to go through the concepts once again.

But having an insight into how these formulas work is just half battle won. Figuring out how to interpret a real-world situation can be quite hard.

Hope you like the Permutation and Combination aptitude questions

P&C is also a very important topic for TCS NQT