Question

exercitiul 5 va rog frumos

Answer 1

combinatii de n cate 2 = n!/ 2!(n-2)!
n!/2!(n-2)!=28
n!= 28*2*(n-2)!
n!= 56(n-2)!
n!/(n-2)!=56
=>
(n-1)*n= 56
=> inmultirea a 2 numere consecutive = 56
56= 7*8
=> n=8
fa si verificarea
8!/2!(8-2)!=28
8!/2*6!= 28
7*8/2=28

Answer 2

$C^2_{n}=28,deci: \frac{n!}{2!(n-2)!}=28,$
simplificand cu (n-2)! si elimiunand numitorul ⇒ n(n-1)=56=7*8, deci n=8