Question 143073
I'm only going to do the first problem here, resubmit the rest and other tutors may assist further.

24.b) (n+2)! / n! = 56 

(n+2)!=(n+2)(n+1)n!

Thus, (n+2)(n+1)=56
n^2+3n+2=56
n^2+3n-54=0
(n+9)(n-6)=0
Which implies n=-9 or n=6. We exclude n=-9 because n! has n € N.

n=6.

Check:
8!/6!=8*7=56.