Verified answer

a) (n - 1)!/(n - 3)! = 90

(n - 1)*(n - 2)*(n - 3)!/(n - 3)! = 90

n² - 3n + 2 = 90

n² - 3n - 88 = 0

(n - 11)*(n + 3) = 0

n = 11

(n,1) - (n,2) = 35

n!/(n-1)! - n!/2!(n-2)! = 35

n*(n-1)!/(n-1)! - n*(n-1)(n-2)!/2*(n-2)! = 35

n - n*(n - 1)/2 = 35

2n - n² + n = 70

n² - 3n + 70 = 0

(n - 10)*(n + 7) = 0

n = 10

pronto

Ui rapaz, fatorial... Vou resolver so a A

(n-3)!/(n-1)!=1/90

(n-3.-2.-1)!/(n-1)!=1/90

(n-3.-2)/n=1/90

(n+6)/n=1/90

n=90.n+540

n=90n+540

n-90n=540

-89n=540

n=540/89

n=6,06.