# Permutations and Combinations – Part 4- Combinations

This article is the fourth in a series of 5 articles. Here we will be discussing a pet topic of all competitive examinations – Combinations. I would advise you to go ahead with this article only after you are clear on:

You will be better equipped to move on with Combinations after you are thorough with the above topics. We will discuss various cases of Grouping & Distribution in the article following this.

All right, let us get started!

## Difference between Permutations and Combinations

Suppose there are 3 bags (A,B and C) in my home and I want to select any 2 out of them to take with me on my holiday. In how many ways can I make the selection?

Clearly I select either AB or BC or AC i.e. 3 ways.

An important point to note is that we are talking about selection and not order here. Obviously whether I select AB or BA makes no difference.

Let us take one more example.

Suppose from a class of 10 students I have to select 3 students for a play, it is a case of Combinations. But, if I have to arrange 3 students in a line from a class of 10 students, it is a case of Permutations.

I hope I have made it clear that in permutations (rearrangement) order matters but in combinations (selections) order does not matter.

We are now in a position to define Combinations.

## Definition of Combinations

Combinations is the selection of some or all of a total of n number of things.

If out of n things we have to select r things (1≤r≤n), then the number of combinations is denoted by ^{n}C_{r} = n!/r!(n-r)! |

Combinations does not deal with the arrangements of the selected things. This explains division by r! which denotes the arrangement of the selected r things.

## Important relation between Permutations and Combinations

‘r’ selected things can be arranged in r! ways.

So, r! x ^{n}C_{r} = ^{n}P_{r}

or, ^{n}C_{r} = ^{n}P_{r} / r!

or^{ n}C_{r} = n! / r! (n-r)!

**Example:** In a class there are 6 boys and 5 girls. In how many ways can a committee of 2 boys and 2 girls be formed?

**Solution:** 2 boys can be selected out of 6 in ^{6}C_{2} ways.

2 girls can be selected out of 5 in ^{5}C_{2} ways.

So the selection can be made in ^{6}C_{2} x ^{5}C_{2} ways. (Product Rule: ‘and’ stands for multiplication)

**Example:** In a class there are 6 boys and 5 girls. A committee of 4 is to be selected such that it contains at least 1 boy and 1 girl.

**Solution:** There are 3 different possibilities now-

i. 1 boy and 3 girls

ii. 2 boys and 2 girls

iii. 3 boys and 1 girl

In the 1st possibility, total number of combinations = ^{6}C_{1} x ^{5}C_{3}

In the 2nd possibility, total number of combinations = ^{6}C_{2} x ^{5}C_{2}

In the 3rd possibility, total number of combinations = ^{6}C_{3} x ^{5}C_{1}

But only one of the above possibilities will occur; 1st OR 2nd OR 3rd.

So the total number of required combinations is ^{6}C_{1} x ^{5}C_{3 }+ ^{6}C_{2} x ^{5}C_{2 }+ ^{6}C_{3} x ^{5}C_{1}

## Some Important Results on Combinations

^{n}C_{r}=^{n}C_{n-r }(0≤r ≤n)^{n}C_{0}=^{n}C_{n}= 1^{n}C_{r}+^{n}C_{r-1}=^{n+1}C_{r}- If
^{n}C_{p}=^{n}C_{q}, then p = q or p + q = n (p,q € W)

## Restricted Combination

The number of combinations of ‘n’ different things taken ‘r’ at a time subject to restriction that p particular things -

i) will never occur = ^{n-p}C_{r}

ii) will always occur = ^{n-p}C_{r-p}

Number of ways of selecting one or more things from a group of n distinct things = ^{n}C_{1} + ^{n}C_{2} + ^{n}C_{3} + …… + ^{n}C_{n} = 2^{n} – 1 .

Number of ways of selecting zero or more things from a group of n distinct things = ^{n}C_{0} + ^{n}C_{1} + ^{n}C_{2} + ^{n}C_{3} + …… + ^{n}C_{n} = 2^{n}

With this we conclude the topic Combinations. Feel free to post interesting questions which we all can share and solve together. In the following article we will discuss the rules on Grouping & Distribution- a very important concept of permutation and combination.