0
$\begingroup$

I try to understand some logic proof and I stumbled upon following:

$$\frac{(N+2m)!}{N!\ m!\ m!} < \frac{({N+2m})^{2m}}{m!\ m!}$$

I don't know why it's true.

This is from a book called "A First Course in Logic" by S.Hedman. I think that it will be really obvious, when I see it, but I wasn't able to work that out.

$\endgroup$
1
$\begingroup$

First, $$n!=n \times (n-1) \times (n-2) \times \cdots \times 1< n \times n \times n \times \cdots \times n = n^n$$

Hence $$ \frac{\left(N+2m\right)!}{N!}<\left(N+2m\right)^{2m+N-N}=\left(N+2m\right)^{2m} $$ Noticing that in fact : $$ \frac{\left(N+2m\right)!}{N!}=\left(N+2m\right)\left(N+2m-1\right)\dots\left(N+1\right) $$

$\endgroup$
1
$\begingroup$

You have: $$\begin{aligned} (N+2m)! &= 1 \times 2 \times 3 \times \dots \times N \times (N+1) \times (N+2) \times \dots (N+2m)\\ &= N! \times (N+1) \times (N+2) \times \dots \times (N+2m)\\ &< N! \times (N+2m)^{2m} \end{aligned}$$

Because for $1 < k \le 2m$, $N+k < N+2m$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.