Problem Statement:
A company has 10 employees, and they want to form a committee of 4 members. However, due to internal dynamics, two specific employees, Alice and Bob, refuse to serve on the committee together. How many different committees can be formed under this condition?
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Solution:
To solve this problem, we need to calculate the total number of possible committees and then subtract the number of committees where both Alice and Bob are included.
#### Step 1: Total number of committees without any restrictions
The total number of ways to choose 4 members out of 10 employees is given by the combination formula:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
Here, \( n = 10 \) and \( r = 4 \). So,
\[
\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4! \cdot 6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210
\]
Thus, there are 210 possible committees if there were no restrictions.
#### Step 2: Number of committees where both Alice and Bob are included
If both Alice and Bob are on the committee, we need to choose the remaining 2 members from the other 8 employees (since Alice and Bob are already chosen). The number of ways to do this is:
\[
\binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8!}{2! \cdot 6!} = \frac{8 \times 7}{2 \times 1} = 28
\]
So, there are 28 committees where both Alice and Bob are included.
#### Step 3: Subtract the restricted committees from the total
To find the number of valid committees where Alice and Bob are not both included, we subtract the number of committees where both Alice and Bob are included from the total number of committees:
\[
\text{Valid committees} = \text{Total committees} - \text{Committees with both Alice and Bob}
\]
\[
\text{Valid committees} = 210 - 28 = 182
\]
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Final Answer:
\[
\boxed{182}
\]
Parent Tip: Review the logic above to help your child master the concept of arithmetic sequence worksheet algebra 1.
