I will do one. Take the greater number, and expand it. for instance, the first
(n-3)!/(n*(n-1)(n-2)(n-3)!)
the (n-3)! terms divide out..
1/(n)(n-1)(n-2)
write each expression without using the factorial symbol.
1)(n-3)!
-----
n!
2) (n+4)!
-------
(n+2)!
2 answers
then the second one would be
(n+4)!/((n+1)(n+2)(n+3)(n+4)(n+2)!)
=
1/(n+1)(n+2)^2 (n+3) ?
(n+4)!/((n+1)(n+2)(n+3)(n+4)(n+2)!)
=
1/(n+1)(n+2)^2 (n+3) ?