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.Com
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) (Show Source): You can put this solution on YOUR website!
Evaluate:
Substitute: and
=
Answer by stanbon(75887) (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.
RELATED QUESTIONS
Solve via direct proof:
n^5 - 64n^3 - n^2 ∈ Θ (n^5)
I'm not exactly sure... (answered by ikleyn)
Find the first five (5) terms of the sequence (begin with n = 1).
A=n!/n^2
I still do (answered by rapaljer,stanbon)
I do not understand the process of evaluating expressions with fractions. I am trying to... (answered by josmiceli)
I have tried this problem for rational equations and I am unable to get the "work shown"... (answered by scott8148)
which is greater?
(n+2)(n+1)(n)(n-1)(n-2) or... (answered by Alan3354)
help me please i dont understand how to do this: {{{5^(3*n+6) =... (answered by Nate,stanbon)
I've been working through factoring polynomials and have solved the following as such;... (answered by stanbon)
I am working with recursive functions. I do not understand the examples given in the... (answered by stanbon)
pls show how is... (answered by stanbon)