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At a game show, there are 7 people (including you and your friend) in the front row.

The host randomly chooses 3 people from the front row to be contestants. The order in which they are chosen does not matter.

How many ways can you and your friend both be chosen?

A. [tex]${ }_5 C_1=5[/tex]
B. [tex]$5 P_3=210[/tex]
C. [tex]$7 P_2=42[/tex]
D. [tex]$7 C_2=21[/tex]


Sagot :

To solve this problem, we need to calculate the number of ways to choose 3 people from a group of 7 people, with the condition that you and your friend must be among those 3 people chosen.

Let's break down the problem step by step:

1. Total People:
- There are 7 people in total, including you and your friend.

2. People to Be Chosen:
- The host needs to choose 3 people from the 7 people in the front row.

3. Condition:
- You (Person A) and your friend (Person B) must be among the 3 chosen people.

Since you and your friend must be chosen, that leaves us with 1 more spot to fill from the remaining people.

4. Remaining People:
- After selecting you and your friend, there are [tex]\(7 - 2 = 5\)[/tex] people remaining.

5. Remaining Spot:
- We need to choose 1 more person from these 5 remaining people to fill the last spot among the 3 people being chosen.

The number of ways to choose 1 person from 5 people can be represented using the combination formula:
[tex]\[ \binom{5}{1} \][/tex]
The combination formula [tex]\(\binom{n}{k}\)[/tex] represents the number of ways to choose [tex]\(k\)[/tex] items from [tex]\(n\)[/tex] items and is defined as:
[tex]\[ \binom{n}{k} = \frac{n!}{k!(n - k)!} \][/tex]

For our case:
[tex]\[ \binom{5}{1} = \frac{5!}{1!(5-1)!} = \frac{5!}{1! \cdot 4!} = \frac{5 \cdot 4!}{1 \cdot 4!} = 5 \][/tex]

Thus, there are 5 ways to choose the 3rd person from the remaining 5 people after you and your friend have been chosen.

Therefore, the number of ways for you and your friend to both be chosen as contestants is:
[tex]\[ \boxed{5} \][/tex]

Answer choice A ([tex]\( \binom{5}{1} = 5 \)[/tex]) is correct.