**Combinations**

Hello Readers

Today’s topic is Combinations (Selection, committee, group)

Each of the different groups or selections which can be formed by taking some or all of a number of objects called combinations.

Suppose we want to select two out of three persons A, B, C. Then possible selections are AB, BC, CA

**Note – **AB and BA represent the same selection.

Suppose we want to select three out of three persons A, B, C. Then possible selections are ABC.

**Note – **ABC and BAC and CAB are the same selection.

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**Number of Combinations – **

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**Note – **

If (n=r), then ^{n}C_{n} = 1 and ^{n}C_{o }= 1

**Important Points:**

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^{n}C_{r }= ^{n}C_{(n-r) }

e.g.

^{16}C_{13} = ^{16}C_{16-13 }= ^{16}C_{3}

Let’s discuss some examples –

**Example 1**

Find the value of ^{5}C_{2}

**Solution – **

**Example 2**

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Find the value of n when ^{n}C_{2} = 105?

**Solution – **

^{n}C_{2} = 105

**Example 3**

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There are 15 persons in a group. They hand shake to each other. Find the different number of handshake.

**Solution – **

**Example – 4**

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From a group of 10 men and 5 women. 4 persons are to be selected to form a committee. Find the different number of ways of selection.

**Solution – **

**Example 5**

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From the group of 10 men and 5 men. 4 persons to be selected such that 3 men and 1 woman to be selected to form a committee. Find the different number of ways of selection.

**Solution –**

**Example 6**

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From a group of 10 men and 5 women. 4 persons are to be selected such that 4 men or 4 women in the group. Find the different number of ways.

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**Solution –**

**Example 7**

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A committee of 5 members is to be formed out of 4 men and 5 women.

(i) In how many ways can a committee consisting of at least 1 woman be formed?

**Solution – **

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^{5}C_{1 }× ^{4}C_{4 }+ ^{5}C_{2 }× ^{4}C_{3 }+ ^{5}C_{3 }× ^{4}C_{2 }+ ^{5}C_{4 }× ^{4}C_{1 }+ ^{5}C_{5 }× ^{4}C_{0}

= 5 × 1 + 10 × 4 + 10 × 6 + 5 × 4 + 1

= 126

(ii) In how many ways can a committee consisting of 3 men and 2 women be formed?

**Solution – **

With this, we will finish this topic.

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