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Calculus and Beyond Homework Help
Counting problem - how many ways....
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[QUOTE="tnich, post: 6048563, member: 639870"] For 1 book on the first shelf, there are ##\binom {15} 1## ways to choose which books are in each group. For the first group there is 1 way to arrange the group, for the second group there are 14! ways. So the number of arrangements for one book on the first shelf is ##\binom {15} 1 (1)(14!) = 15!## Similarly, for 2 books on the first shelf, there are ##\binom {15} 2 (2!)(13!) = 15!## arrangements and for k books on the first shelf, there are ##\binom {15} k (k!)(15-k)! = 15!## arrangements How many possible values of k are there? [/QUOTE]
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Counting problem - how many ways....
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