SOLUTION: n! _______ 2!(n-2)! I understand that 5! is 5*4*3*2*1 I do not understand how to figure n! I understand that 2! is 2*1 and (n-2)! is (n-2)*(n-1) which would make the probl

Algebra ->  Permutations -> SOLUTION: n! _______ 2!(n-2)! I understand that 5! is 5*4*3*2*1 I do not understand how to figure n! I understand that 2! is 2*1 and (n-2)! is (n-2)*(n-1) which would make the probl      Log On


   

Question 174238: n!
_______
2!(n-2)!
I understand that 5! is 5*4*3*2*1
I do not understand how to figure n!
I understand that 2! is 2*1 and (n-2)! is (n-2)*(n-1) which would make the problem
n!
________________
2*1(n-2)*(n-1)
Atleast I think that's what the problem would be, but now what? What do I do with the n! ??? I'm so confused. PLEASE HELP!! Thank you in advance!
Found 2 solutions by Earlsdon, stanbon:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Evaluate:
%28n%21%29%2F2%21%28n-2%29%21 Substitute: n%21+=+n%2A%28n-1%29%2A%28n-2%29%21 and 2%21+=+2
n%2A%28n-1%29cross%28%28n-2%29%21%29%2F%282%2Across%28%28n-2%29%21%29%29=highlight%28n%28n-1%29%2F2%29

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
n!
_______
2!(n-2)!
--------------------

= [n(n-1)(n-2)!] / [2!*(n-2)!]
Cancel the (n-2)! that is common to the numerator and the
denominator to get:
= [n(n-1)]/2
=================
Cheers,
Stan H.