The concept of combination is used:

(a) to find number of ways element can be selected from given set, and

(b) when order is not important

**Example 01**

Find number of ways we can select two objects from given objects A, B, C, D

**Manual Calculation**

By solving the question manually we get:**AB, AC, AD, BC, BD, CD**

These are the possible combinations (Total 6)**Note:**

We have not included BA, CB etc. because AB = BA (both are same object & order is not important)

**Solving using Combination concept**

Without manual counting, use following formula:

C (n, r) = C (4, 2) = \frac{4!}{( 4-2) !\times 2!}\ **C (4, 2) = 6**

Same question easily solved using combination formula

**Example 02**

Three Badminton players are available Adam, Bill, Clark. You have to select two of them to form a team. Find the different possible combination**Manual Counting****(Adam, Bill), (Bill, Clark), (Clark, Adam)**

There are three possible combination

Note: we have ignored elements like (Bill, Adam) because this will lead to double counting as:

(Adam, Bill) = (Bill, Adam)

Here order is not important

**Solving using Combination formula**

C (n, r) = C(3, 2) = \frac{3!}{( 3-2) !\times 2!}\

C (3, 2) = 3

Hence there are three possible combinations

**Combination Definition and Formula**

Combination is a math concept through which you can find the number of possible selection of element from a given set, when the order of element is not important.

**Symbol & Formula of Combination**

Where;

n = number of element in set

r = selected elements

**Difference between Permutation & Combination**

Permutation is used when order of element is important.

For example, in question where arrangement AB & BA are considered different, permutation should be used.

Combination is used when order is not important

When arrangement AB & BA are considered same, combination can be used.

**Example 01**

Find number of ways you can choose two elements from the set {A, B, C, D}.

Here order of elements is important [ i.e AB & BA are different]**Solution using Permutation **

P (4, 2) = \frac{4!}{( 4-2) !}

P (4, 2) = 12

Hence there are 12 possible combinations.**Permutation Analysis**

Below image shows how permutation works

The Permutation Formula:

1. First select the elements

2. Then reshuffle the element to gather all possible combination

3. Repeat the step 1

After all the steps we get 12 total elements as our answer

**Solution using combination**

Combination formula find the number of possible combination without order sequence

C (4, 2) = \frac{4!}{( 4-2) !\ 2!} **C (4, 2) = 6**

Now we have to shuffle the two selected elements

=> 6 * 2!

=> **12**

We got the same answer using combination formula.

Let us understand how the above process works through below illustration

**Solution Explanation using Combination**

1. C (4,2) gave all possible selection

AB, AC, AD, BC, CD, BD

2. The selected elements should be reshuffled to get all the possible options

AB reshuffled to give AB, BA

AC ————————-> **AC, CA**

AD ————————-> **AD, BA**

BC ————————-> **AB, BA**

CD ————————-> **AB, BA**

BD ————————-> **AB, BA**

For reshuffling two elements we multiplied **2!**

3. Final Equation was => **c (4,2) * 2!**

**Relationship between Permutation and Combination**

In the above question we got the same answer using different technique, so there must be common link between the concept of permutation and combination.**In Permutation Method we used following equation**

P (4, 2) = \frac{4!}{( 4-2) !}

**In Combination Method, we used following method**

=> **C (4, 2) * 2!**

=> \frac{4!}{( 4-2) !\ 2!} * 2!

Equating both the equation, we will get following relationship**P (n, r) = C (n, r) * r!**

**Note**:

Just remember the following concept:

(a) Combination [**C(n, r)**] is only selection

(b) Permutation [**P(n,r)**] is both selection and rearrangement

**Important Properties of Combinations**

1.** When r = n, the value of combination will be 1**

2. **When r = 0, the value of combination will be 1**

3. In a collection of n objects, selecting **r** objects is same as rejecting **(n – r)** objects

4. **Combination Formula**

5. **Combination Equality Concept**

The above two combinations are equal in two condition

(1) Either **a = b**, or,

(2) ** n = a + b**

All the above mentioned concepts are important.

They will help you solve worksheet problems from this chapter.

**Solved Combination Problems**

**Question 01**

A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?

**Solution Part 01**

Total Men = 2

Total Women = 3

Total People = 2 + 3 = 5

Let us represent all the 5 people as M1, M2, W1, W2, W3

Number of ways 3 people can be selected = C( 5, 3) = \frac{5!}{3!\ 2!}\

==> C (5, 3) = 10

Hence there are 10 ways in which three person can be selected.

**Solution Part 02**

Selection of 1 Man and 2 Women

Hence there are 6 possible ways in which the committee will be formed where there will be 1 Man and 2 Woman

**Question 02**

How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE ?

**Solution**

Let us first separate vowels and consonant from the word INVOLUTE

Vowels: **I, O, U, E** [ 4 choices]

Consonant: **N, V, L, T** [ 4 choices]

**On solving the above expression we will get = 2880 possibilities**