# Combination question?

I have question it is a practise question for my textbook.

You are given $5$ books and $7$ bookshelves. How many ways are there to place these books on the shelves? (Order matters). I checked the back of the book and its says $\dfrac{11!}{6!}$.

Anyone care to explain is that?

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Split the process of arrangement in two steps, count the ways you can conduct each step and then use the multiplication principle to find the total number of ways you can conduct the whole process. Here you need two steps.

First step. Choose the number of books that will come in each shelf. That is, determine numbers $x_1,\ldots,x_7$ such that $$x_1+x_2+\ldots+x_7=5$$ subject to $0\le x_i$. You can do that in $$\dbinom{5+7-1}{7-1}=\dbinom{11}{6}$$ ways (see here for an explanation).

Second step.Then order your books in.You can do it in $5!$ ways. But now you already know how many books come in each shelf, so determining the order was the only that was remaining.

By the multiplication principle you have that the total number of ways (i.e. both steps) is $$\dbinom{11}{6}\cdot5!=\frac{11!}{5!6!}\cdot5!=\frac{11!}{6!}$$

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kil- Just an add-on to a great explanation: the problem is a variant of the "stars and bars" problem in combinatorics. –  amWhy Feb 27 '14 at 15:48

Put $6$ cardboard dividers on the last shelf, representing the first $6$ shelves. The first book can be placed in $7$ slots and henceforth acts as an additional divider; the second book can then be placed in $8$ slots, and becomes a divider too, and so on. In all we have $7\cdot8\cdot\ldots\cdot 11={11!\over 6!}$ choices.

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The answer by Stefanos is better, since it puts together two natural ingredients. There are other ways. Imagine instead putting the books on a single shelf, together with $6$ books dividers. (Then for the real shelving, we will put the books up to the first divider onto Shelf $1$, and the books from the first divider to the second onto shelf $2$, and so on.)

Imagine that the books are called $a,b,c,d,e$. Write $D$ for a divider. We want to count the number of $11$-letter "words" that use each of $a,b,c,d,e$ once, and $6$ copies of $D$.

If the $Ds$ were distinct, say $D_1$ to $D_6$, there would be $11!$ words. Since they are identical, we divide by $6!$.

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Well first book can be put on any of 7 shelves,second book can be placed 8 ways,any of the shelves and switching place with the first book.Third book can be also put on every shelf + it can be switched with the first two books and etc. so the result is $7*8*9*10*11$

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