# Show that $(n-j)! \leq \frac{n!}{(j+1)!}$

I have to show that $$(n-j)! \leq \frac{n!}{(j+1)!}$$

I tried to develop $(n-j!)$ as $\frac{n!}{n(n-1)\cdot\cdot\cdot(n-j+1)}$ but I have no clue after that because I end up with $$\frac{1}{n(n-1)\cdot\cdot\cdot(n-j+1)} \leq \frac{1}{(j+1)!}$$

$${n(n-1)\cdot\cdot\cdot(n-j+1)} \geq {(j+1)!}$$

• It's not true for $j=n$. – Thomas Andrews Oct 18 '17 at 18:09

You could also just induct on $n$. The base case for $n = j+1$ is easy. The induction step isn't difficult either, as $$(n+1-j)! = (n+1-j) (n-j)! \leq (n+1-j) \frac{n!}{(j+1)!} \leq (n+1) \frac{n!}{(j+1)!} = \frac{(n+1)!}{(j+1)!}.$$
If $j\geq 0$ and $j<n$ then $n+1 \leq {n+1\choose j+1}$, thus
$$1\leq {1\over n+1}{n+1\choose j+1} = {n! \over (j+1)!(n-j)!}$$
• It might also be useful to not that $n+1=\binom{n+1}{1}$, to make the first inequality more obvious. – Thomas Andrews Oct 18 '17 at 18:09
• why $\binom{n+1}{1} \leq \binom{n+1}{j+1}$ – Rom Oct 18 '17 at 18:11
• Let $F$ be a family of all $j$ element subset of set $M=\{1,2,....,n\}$. For each element in $F$ let $m$ be minimum and $M$ maximum of this element. Now we have a surjective map $f:F\to M$ which takes $X$ to $m$ if $m<j$ and other vise to M. Thus $|M|\geq |F|$. – Aqua Oct 18 '17 at 18:26
• That's why must be $j<n$ ? – Rom Oct 18 '17 at 18:35