n!/(n-2)!=90
3 answers
n=10
how did u get the n=10
rewrite it as ...
n!/(n-2)!=90
[n(n-1)(n-2)(n-3)..(2)(1)]/(n-2)(n-3)..(2)(1)] = 90
n(n-1) = 90
n^2 - n = 90 = 0
(n-10)(n+9)=0
n=10 or n= -9 , but n > 0
so n=10
n!/(n-2)!=90
[n(n-1)(n-2)(n-3)..(2)(1)]/(n-2)(n-3)..(2)(1)] = 90
n(n-1) = 90
n^2 - n = 90 = 0
(n-10)(n+9)=0
n=10 or n= -9 , but n > 0
so n=10