Question 1022433: Eight prizes are to be given to eight different people in a group of ten. In how many ways can a first prize, a second prize, a third prize and five fourth prizes be given?
Answer by mathmate(429)   (Show Source):
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Question:
Eight prizes are to be given to eight different people in a group of ten. In how many ways can a first prize, a second prize, a third prize and five fourth prizes be given?
Solution:
Start with the first prize, there are 10 choices.
Then the second prize has 9 remaining candidates.
The third prize has 8 remaining people.
Next, we choose five people for the fourth prize out of 7 remaining people in C(7,5)=21 ways, where C(n,r) is the binomial coefficient of choosing r objects out of n, and given by C(n,r)=n!/(r!(n-r)!). Note that order does not count in the choice of the five people for the fourth prize.
The total number of arrangement of prizes is therefore 10*9*8*21=15120 ways.
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