SOLUTION: 4. Four customers are to be chosen for prizes 1,2,3, and 4th place out of a set of 25 different customers. a) How many possible combinations are there for the four winners,

Algebra ->  Probability-and-statistics -> SOLUTION: 4. Four customers are to be chosen for prizes 1,2,3, and 4th place out of a set of 25 different customers. a) How many possible combinations are there for the four winners,       Log On


   



Question 939272: 4. Four customers are to be chosen for prizes 1,2,3, and 4th place out of a set of 25 different customers.
a) How many possible combinations are there for the four winners, in which all we care about is whether the person was chosen at all, not in what position they finished?
b) How many different orderings of the top 4 would be possible (order does matter)?

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
4. Four customers are to be chosen for prizes 1,2,3, and 4th place out of a set of 25 different customers.
a) How many possible combinations are there for the four winners, in which all we care about is whether the person was chosen at all, not in what position they finished?
Ans: 25C4 = (25*24*23*22)/(1*2*3*4) = 12650
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b) How many different orderings of the top 4 would be possible (order does matter)?
Ans: 25P4 = 25*24*23*22 = 303600
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Cheers,
Stan H.
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