Respuesta :

LammettHash
Of [tex]n[/tex] elements, there are [tex]{}_nC_r=\dfrac{n!}{r!(n-r)!}[/tex] ways of choosing any [tex]r[/tex] elements. So the number of subsets that can be chosen from the set of 13 elements, each consisting of 4 to 8 elements, is

[tex]\displaystyle\sum_{r=4}^8{}_{13}C_r={}_{13}C_4+\cdots+{}_{13}C_8=6721[/tex]

To compute the actual numbers, you have, for example,

[tex]{}_{13}C_4=\dfrac{13!}{4!(13-4)!}=\dfrac{13!}{4!9!}[/tex]
[tex]=\dfrac{13\times12\times\cdots\times6\times5}{9\times8\times\cdots\times2\times1}[/tex]
[tex]=\dfrac{13\times12\times11\times10}{4\times3\times2\times1}[/tex]
[tex]=13\times11\times5[/tex]
[tex]=715[/tex]

so there are 715 ways of picking subsets of size 4. Compute the others similarly, then add them up.

Otras preguntas