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I used stars and bars for this, and my reasoning states that using 2 bars and 6 stars would work. There are 5 spaces to put the bars, so it is 5 choose 2, which is 10, but the answer is much larger at 540.
Can I solve this using stars and bars, and how can I do this with PIE?

Gerard L.
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    You have correctly answered the question in the case where the books are indistinguishable, but here the books are distinguishable. – angryavian Oct 28 '17 at 19:46
  • Maybe worth noting that, in general, the number of surjections from ${1,\cdots, n}$ to ${1,\cdots, m}$ with $n>m$ is not trivially determined. Of course, in this case $n,m$ are quite small. In general they are given by Stirling Numbers of the Second Kind – lulu Oct 28 '17 at 19:48
  • "Stars and bars"? PIE? How about thinking.Call the three people "A", "B", and "C" in any order you wish. Choose one book and give it to A. There are 6 choices. Choose another book and give it to B. There are 5 choices. Choose a third book and give it to C. There are 4 choices. Now there are three books you can give to the three people in any order you wish. There are 3*3= 9 ways to do that. – user247327 Oct 28 '17 at 19:53
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    @user247327 I believe you have over-counted by quite a bit. If person A receives multiple books, you have separately counted situations where she receives "A Tale of Two Cities" as part of the "first three books" stage or as part of the "remaining three books" stage, even if person A ends up with the same set of books. – angryavian Oct 28 '17 at 20:39

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