Let's solve this step-by-step.
First, identify the problem:
We need to calculate how many ways we can choose 3 people out of the 8 in the front row, such that you and your friend are among the 3 chosen.
### Step 1: Total Ways to Choose 3 People from 8
We start by calculating the total number of ways to choose any 3 people out of the 8 in the front row. This is given by the combination formula, where order does not matter:
[tex]\[
\binom{8}{3} = \frac{8!}{3!(8-3)!}
\][/tex]
This calculation simplifies to:
[tex]\[
\binom{8}{3} = \frac{8!}{3!5!} = 56
\][/tex]
So, there are 56 ways to choose any 3 people from the 8. Therefore, option C: [tex]\(\binom{8}{3} = 56\)[/tex] is correct for the total number of ways to choose 3 people from 8.
### Step 2: Ways to Choose 1 More Person (Given You and Your Friend are Chosen)
Since you and your friend are to be chosen, we only need to choose 1 more person from the remaining 6 people. The number of ways to choose 1 person out of 6 is:
[tex]\[
\binom{6}{1} = 6
\][/tex]
Therefore, the number of ways to choose you, your friend, and one more person out of the remaining 6 is 6.
### Conclusion
The number of ways for you and your friend to both be chosen is [tex]\(\binom{6}{1} = 6\)[/tex], which corresponds to option B.
Thus, the correct answer is:
B. [tex]\(\binom{6}{1} = 6\)[/tex]
