Question 1001515
(3n+2)! / (3n-3)! = (3n+2)*(3n+1)*(3n)*(3n-1)*(3n-2)*(3n-3)! divided by (3n-3)!


the (3n-3)! in the numerator and denominator cancel out and you are left with:


(3n+2)*(3n+1)*(3n)*(3n-1)*(3n-2)


for example:


let n = 7


3n+2 = 23
3n+1 = 22
3n+0 = 21
3n-1 = 20
3n-2 = 19
3n-3 = 18


(3n+2)! / (3n-3)! = (3n+2)*(3n+1)*(3n)*(3n-1)*(3n-2)*(3n-3)! divided by (3n-3)! which becomes:
23! / 18! = (23 * 22 * 21 * 20 * 19 * 18! divided by 18! which becomes:
(23 * 22 * 21 * 20 * 19)


the result is 4037880


i used my ti-84 to take 23! and divide it by 18! and i got 4037880.


this confirms the solution is correct.