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 June 27th, 2019, 10:30 AM #1 Newbie   Joined: Jan 2019 From: Bosnia and Herzegovina Posts: 9 Thanks: 0 Combinatorics On how many ways Vlad, Alex and Mike can share 5 books, whereby can happen that someone didn't get a single book if books are same? _ (7) The answer is P(5,2) or 7! / (5! * 2!) = 21. I'm not sure where did these numbers came from, so if someone could explain it.
 June 27th, 2019, 10:35 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 I'm afraid I can't understand the translation into English. Can you try to explain the problem conditions again in more detail?
 June 27th, 2019, 10:57 AM #3 Newbie   Joined: Jan 2019 From: Bosnia and Herzegovina Posts: 9 Thanks: 0 So there are 3 different people and there are 5 same books. It can happen that one person gets all 5 books and other two don't get any. Question is on how many ways they can share those books. Hope it's more understandable now. This is answer: https://prnt.sc/o7lbvq Last edited by qu4lizz; June 27th, 2019 at 11:00 AM.
June 27th, 2019, 12:12 PM   #4
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Quote:
 Originally Posted by qu4lizz So there are 3 different people and there are 5 same books. It can happen that one person gets all 5 books and other two don't get any. Question is on how many ways they can share those books. Hope it's more understandable now. This is answer: https://prnt.sc/o7lbvq
This is just what's known as a "stars and bars" problem.

Sorting, in this case, 5 identical balls (books), into 3 distinct bins (people).

The general formula for sorting $n$ identical balls into $k$ distinct bins is

$N = \dbinom{n+k-1}{k-1}$

You can Youtube "stars and bars" to see a detailed explanation of why.

Here $n=5,~k=3$ so $N = \dbinom{5+3-1}{3-1} = \dbinom{7}{2} = 21$

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