In how many ways can a committee of four be selected from nine so as to always include a particular man?

Options

A. 48
B. 56
C. 70
D. 126

Explanation

To solve the problem of selecting a committee of four from nine people while ensuring that a particular man is always included, we can follow these steps:

### Step 1: Understand the Problem
We need to form a committee of four members from a total of nine people, but one specific person must always be included in the committee.

### Step 2: Fix the Particular Man
Since one specific man must always be included in the committee, we can consider him as already chosen. This means we only need to select the remaining members of the committee.

### Step 3: Determine Remaining Members
After including the particular man, we are left with eight people (since one has already been chosen). We need to select three more members from these eight people to complete the committee of four.

### Step 4: Use the Combination Formula
The number of ways to choose \( r \) members from \( n \) members is given by the combination formula:
\[ nCr = \frac{n!}{r!(n-r)!} \]
In our case, we need to calculate \( 8C3 \) because we are choosing 3 members from the remaining 8.

### Step 5: Calculate \( 8C3 \)
Using the combination formula:
\[ 8C3 = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \cdot 5!} \]
Now, we can simplify this:
\[ 8C3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \]
Calculating the numerator:
\[ 8 \times 7 \times 6 = 336 \]
Calculating the denominator:
\[ 3 \times 2 \times 1 = 6 \]
Now, divide the numerator by the denominator:
\[ 8C3 = \frac{336}{6} = 56 \]

### Step 6: Conclusion
Thus, the total number of ways to select a committee of four from nine people, ensuring that the particular man is always included, is **56**.

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