A committee of five members is to be formed out of Males , 6 Females and 3 children ,In how many different ways can it be done if -?
The committee should include all the 3 childs ?

Options

A. 90
B. 180
C. 21
D. 55

Explanation

To solve the problem of forming a committee of five members that includes all three children, we can follow these steps:

### Step 1: Understand the Composition of the Committee
We need to form a committee of 5 members from a total of 5 males, 6 females, and 3 children. The committee must include all 3 children.

### Step 2: Select the Children
Since the committee must include all 3 children, we select all of them. There is only 1 way to do this, as we are taking all 3 children.

**Hint:** When you are required to include all members of a specific group, the selection from that group is fixed.

### Step 3: Determine Remaining Members
After selecting the 3 children, we need to fill the remaining 2 positions of the committee. The remaining members can be selected from the 5 males and 6 females.

### Step 4: Calculate the Total Available Members
The total number of available members (males + females) is:
- Males: 5
- Females: 6
- Total = 5 + 6 = 11 members

### Step 5: Choose 2 Members from 11
Now we need to choose 2 members from the 11 available members. The number of ways to choose 2 members from 11 is given by the combination formula:

\[ \text{Number of ways} = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

Where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.

In our case, \( n = 11 \) and \( r = 2 \):

\[ \binom{11}{2} = \frac{11!}{2!(11-2)!} = \frac{11!}{2! \cdot 9!} \]

### Step 6: Simplify the Combination
We can simplify this as follows:

\[ \binom{11}{2} = \frac{11 \times 10}{2 \times 1} = \frac{110}{2} = 55 \]

### Step 7: Conclusion
Thus, the total number of different ways to form the committee of 5 members, including all 3 children, is **55**.

**Final Answer:** 55 different ways.

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