# Inequality involving two unknowns and factorials

1. Jul 7, 2007

### Rosey24

I need to prove the following:
n!/r! >= r^(n-r)

With r and n as natural numbers and n>=r

I know the LHS will end up being (n-r) terms long as the first r! will cancel out of n! (n>=r), but as they're both unknown, I just left it as
1*2*3*...*(n-2)*(n-1)*n

1*2*3*...*(r-2)*(r-1)*r

and I looked at the RHS as r^n/r^r, but I'm not sure how to work with these two sides. :uhh:

Any tips on different ways to approach this would be greatly appreciated.

Thanks!

Update: Is it enough to do the following, using induction:
Prove that for n=1, it works as 1 >= 1
and then, assuming the inequality holds for n, try n+1:
(n+1)!/r! >= r^[(n+1)-r]
which is just
n!(n+1)/r! >= r^(n-r)*r
and, as n>=r, (n+1)>=r, so, by induction, it's holds for all n.

Last edited: Jul 8, 2007
2. Jul 8, 2007

### HallsofIvy

Staff Emeritus
That's the way I would do it! For a moment I thought- "He's assuming r= 1 and he can't do that", but you can. Since r is a natural number, less than or equal to n, if n= 1 then r= 1.

3. Jul 8, 2007

### huyen_vyvy

LHS= (r+1)*(r+2)...*n , there are n-r terms and each term is greater than r . Hence, the LHS is greater than r^(n-r), right?

4. Jul 8, 2007

### Gib Z

Ahh I like the way this new guy thinks :) Much simpler than induction, original and just plain :) which is good lol.

5. Jul 8, 2007

### Rosey24

Thanks to everyone who responded!

I appreciate the simplicity of your response, huyen_vyvy.

One small note: it's a she who assumed r=1, not a he.