Of \(n\) elements, there are \({}_nC_r=\dfrac{n!}{r!(n-r)!}\) ways of choosing any \(r\) elements. So the number of subsets that can be chosen from the set of 13 elements, each consisting of 4 to 8 elements, is
\(\displaystyle\sum_{r=4}^8{}_{13}C_r={}_{13}C_4+\cdots+{}_{13}C_8=6721\)
To compute the actual numbers, you have, for example,
\({}_{13}C_4=\dfrac{13!}{4!(13-4)!}=\dfrac{13!}{4!9!}\)
\(=\dfrac{13\times12\times\cdots\times6\times5}{9\times8\times\cdots\times2\times1}\)
\(=\dfrac{13\times12\times11\times10}{4\times3\times2\times1}\)
\(=13\times11\times5\)
\(=715\)
so there are 715 ways of picking subsets of size 4. Compute the others similarly, then add them up.
