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So there are $K$ different books and $L$ copies of each in the library. The number of way's in which I can make a selection of one or more books is?

Deiknymi
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1 Answers1

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Assume you never want to choose more than one copies of any book.

If you select one of the books only, you have $KL$ ways to select the copies. If you select two of the books, you have $\binom{K}{2}L^2$ ways to select the copies. Similarly, if you select $i$ books, you have $\binom{K}{i}L^i$ ways to select the copies, for $i$ from $1$ up to $K$.

Summing all ways together, the number of ways is $$\begin{align} \sum^K_{i=1}\binom{K}{i}L^i =& \sum^K_{i=0}\binom{K}{i}L^i - 1\\ =& \sum^K_{i=0}\binom{K}{i}L^i1^{K-i} - 1\\ =&\left(L+1\right)^K-1 \end{align}$$

You can also think there is an extra dummy copy for each book. The problem becomes, for each of the $K$ books, you either select one of the $L$ copies, or you skip choosing that book (hence the extra dummy copy). The total number of ways, minus the choice of skipping all books, is the answer above.

peterwhy
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  • sir can you please explain the $\binom{K}{2}L^2$ a little bit please – Deiknymi Sep 15 '13 at 01:48
  • From the library of $K$ different books, you have $\binom{K}{2}$ ways to choose two books. For each of the two books, there are $L$ ways to select a copy. – peterwhy Sep 18 '13 at 10:33