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.8C3 = 56B.6C1 = 6C.6P2 = 30D.8P3 = 336SUBMITarrow_backPREVIOUS
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.8C3 = 56B.6C1 = 6C.6P2 = 30D.8P3 = 336SUBMITarrow_backPREVIOUS
Solution
The problem is asking for the number of ways you and your friend can both be chosen out of 3 people from a group of 8.
Since you and your friend are already chosen, we are left with choosing 1 person from the remaining 6 (8 total - you - your friend).
This is a combination problem because the order in which the people are chosen does not matter.
The formula for combinations is C(n, r) = n! / [r!(n - r)!], where n is the total number of options, r is the number of options chosen, and "!" denotes factorial.
So, we need to find C(6, 1) which equals 6! / [1!(6 - 1)!] = 6.
Therefore, the answer is B. 6C1 = 6.
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