- #1
Rosey24
- 12
- 0
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.
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: