3(n+1)!/(n-1)=126

Question

This is practicing questions and I'm stumped with what to do with the 3 in the front.

Answer 1

Divide through by 3 and expand: (n+1)n(n-1)!/(n-1)!=42; (n+1)n=42; n^2+n-42=0, (n+7)(n-6)=0; n=6.

CHECK: 3(7!)/5!=3*5040/120=3*504/12=504/4=126.

However, the question says the denominator is n-1 not (n-1)!. Therefore the question changes to:

(n+1)n(n-2)!=42. The lowest value for n is 2: 3*2=6<42; put n=3: 4*3=12<42; put n=4: 5*4*2=40<42; but when n=5: 6*5*6=180>42. Therefore, there is no solution, and the revision of the question so that (n-1)! is the denominator is the interpretation leading to a solution.