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How can you divide 5 different books among 3 children such that each child gets at least one book? (no book can be divided)

Some one please tell me how to solve this.

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Inclusion-exclusion principle – Noturab Mar 31 '13 at 22:52

This is the number of onto functions from a $5$-element set to a $3$-element set. Here, principle of inclusion & exclusion is useful.

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I would use inclusion-exclusion. Count the total ways to distribute the books. Subtract three times the number of ways to give them to a specific pair. But you have subtracted the ways to gove them all to one child twice (paired once with each of the others) so add them in once.

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use the stars and bars method to see that there are 6 ways to give out 5 identical books.

Of these there are 3 combinations where you give 3 to one kid and 1 to the others and there are 3 combinations where you give 2 to one kid and 1 to the other.

in the ones where you give three to one kid divide by 3! to get the answer of 3*5!/3!=60 and in the one where you give 2 to 2 kids divide by 2(2!) to get 3*5!/2*2!= 90 so therefore there are 150 ways to do it.

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