At a game show, there are 8 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.6P2 = 30B.8C3 = 56C.6C1 = 6D.8P3 = 336
Question
At a game show, there are 8 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.6P2 = 30B.8C3 = 56C.6C1 = 6D.8P3 = 336
Solution
The problem is asking for the number of ways you and your friend can both be chosen out of 8 people, with 3 people being chosen in total.
Since you and your friend are already chosen, we are left with choosing 1 person out of the remaining 6 (8 total - you - your friend).
This is a combination problem (order does not matter), so we use the combination formula nCr = n! / r!(n-r)!.
Here, n = 6 (remaining people) and r = 1 (remaining spot to fill).
So, the answer is 6C1 = 6.
Therefore, the correct answer is C.6C1 = 6.
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