Assuming the question is actually:
(n+7)! / (n!(n+6)(n+4))=9!/3
and n>0.
First simplify the left-hand side by cancelling n!:
(n+7)! / (n!(n+6)(n+4))
=(n+1)*(n+2)*(n+3)*(n+4)*(n+5)*(n+6)*(n+7)/((n+6)*(n+4)
=(n+1)*(n+2)*(n+3)*(n+5)*(n+7)
So the problem reduces to solving the quintic
(n+1)*(n+2)*(n+3)*(n+5)*(n+7)=9!/3
Analytic solution of the quintic is likely to be overly complex.
We can take advantage of the fact that the left hand side is strictly monotonic, and the right hand side is a constant.
Solving by binary search (of integers) gives the solution n=7 in a couple of tries.
Find the value of n (n is an integer).
(n+7)! / n!(n+6)(n+4) = 9!/3
1 answer