Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. – Zev Chonoles Mar 31 '13 at 22:34
Inclusion-exclusion principle – Noturab Mar 31 '13 at 22:52

3 Answers 3

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

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer
I changed the answer – dREaM Mar 31 '13 at 23:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.