Question

Encontre n se1) A(n,2) = 110;2) A (n,n) = 5040

Answer 1

A(n,2) = 110
n!/(n-2)! = 110
[n*(n-1)*(n-2)!] / (n-2)! = 110
n*(n-1) = 110
n²-n-110 = 0
/\ = (-1)² -4*1*(-110)
/\ = 1+440
/\ = 441
n = (-(-1)+/- \/441)/2
n = (1+/- 21)/2
n' = (1+21)/2 --> n' = 22/2 --> n' = 11
n" = (1-21)/2 --> n" < 0 --> não convém

A(n,n) = 5040
n! / (n-n)! = 5040
n! / 0! = 5040
0! = 1
n! / 1 = 5040
n! = 5040
5040|2
2520|2
1260|2
0630|2
0315|3
0105|3
0035|5
0007|7
0001

5040 = 7*(2*3)*5*(2*2)*3*2*1

5040 = 7*6*5*4*3*2*1
5040 = 7! --> n = 7