A delegation of 6 members is to be sent abroad out of 12 members. In how many ways can the selection be made so that , a particular member is included?

Answer

To solve the problem of selecting a delegation of 6 members from a total of 12 members, where a particular member must always be included, we can follow these steps:

### Step 1: Understand the problem
We need to select 6 members from a total of 12 members, with the condition that a specific member (let's call them Member A) is always included in the selection.

### Step 2: Fix the selected member
Since Member A is already included in the delegation, we only need to select the remaining 5 members from the other 11 members (12 total members - 1 fixed member = 11 remaining members).

### Step 3: Calculate the number of ways to choose the remaining members
We need to calculate how many ways we can choose 5 members from these 11 remaining members. This can be done using the combination formula:

\[ nCk = \frac{n!}{k!(n-k)!} \]

In our case, \( n = 11 \) and \( k = 5 \):

\[ 11C5 = \frac{11!}{5!(11-5)!} = \frac{11!}{5! \cdot 6!} \]

### Step 4: Simplify the calculation
We can simplify \( 11C5 \) as follows:

\[ 11C5 = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} \]

### Step 5: Calculate the result
Now, let's calculate the numerator and denominator:

- **Numerator**:
\[ 11 \times 10 = 110 \]
\[ 110 \times 9 = 990 \]
\[ 990 \times 8 = 7920 \]
\[ 7920 \times 7 = 55440 \]

- **Denominator**:
\[ 5 \times 4 = 20 \]
\[ 20 \times 3 = 60 \]
\[ 60 \times 2 = 120 \]
\[ 120 \times 1 = 120 \]

Now, divide the numerator by the denominator:

\[ 11C5 = \frac{55440}{120} = 462 \]

### Final Answer
Thus, the number of ways to select the delegation of 6 members, including the particular member, is **462 ways**.

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