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At a game show, there are 6 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]\( \binom{6}{3} = 20 \)[/tex]

B. [tex]\( 6P_3 = 120 \)[/tex]

C. [tex]\( \binom{4}{1} = 4 \)[/tex]

D. [tex]\( 6P_2 = 30 \)[/tex]


Sagot :

Let's carefully determine the number of ways you and your friend can both be chosen as contestants from the 6 people in the front row.

1. Total people in the front row: 6 people (including you and your friend).

2. Total contestants to be chosen by the host: 3 people.

Given that you and your friend must be chosen among the 3 contestants, only one spot remains to be filled by any one of the remaining people.

3. Remaining people to choose from: After you and your friend are chosen, there are [tex]\(6 - 2 = 4\)[/tex] people left.

4. Ways to choose the remaining contestant: The last contestant can be chosen from these 4 people.

The number of ways to choose 1 person out of the remaining 4 people is given by the combination formula [tex]\(\binom{4}{1}\)[/tex], which calculates the number of ways to choose 1 item from a set of 4 items without regard to order.

[tex]\[\binom{4}{1} = 4\][/tex]

This calculation tells us there are 4 different ways to choose the last contestant among the remaining 4 people once you and your friend have already been chosen.

Therefore, the number of ways you and your friend can both be chosen is:
[tex]\[ \boxed{4} \][/tex]

So the correct answer is:
C. [tex]\(\binom{4}{1} = 4\)[/tex]