SOLUTION: If the police have 8 suspects, how many different ways can they select 5 for a lineup?
The answer that I came up with was 6720 ways. But its wrong. Was i supposed to do this prob
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-> SOLUTION: If the police have 8 suspects, how many different ways can they select 5 for a lineup?
The answer that I came up with was 6720 ways. But its wrong. Was i supposed to do this prob
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Question 1111379: If the police have 8 suspects, how many different ways can they select 5 for a lineup?
The answer that I came up with was 6720 ways. But its wrong. Was i supposed to do this problem as a permutation or a combination? Or neither? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! If the police have 8 suspects, how many different ways can they select 5 for a lineup?
The answer that I came up with was 6720 ways. But its wrong. Was i supposed to do this problem as a permutation or a combination? Or neither?
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You are selecting groups so use a combination formula.
8C5 = 8C3 = (8*7*6)/(1*2*3) = 8*7 = 56
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Cheers,
Stan H.
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