# Strictly decreasing function of k

How do I prove $(_nC_k)$k!$(_mC_1)$(n+m-k-1)!/(m+n)! is a strictly decreasing function of k when m>=2.
The only way I found so far to proceed is to find 1st derivative and show that it is less than 0 for m>=2. Is it the right way to proceed? Is there a better way?

• I suppose it is something like binomial coefficients, but what exactly is $\;_nC_k\;$ ? – DonAntonio Sep 14 '16 at 6:15
• Yes. It is a binomial coefficient. – roang Sep 14 '16 at 6:15

Put

$$a_k:=_nC_k\,k!\,_mC_1\frac{(n+m-k-1)!}{(m+n)!}=\binom nkk!\binom m1\frac{(n+m-k-1)!}{(m+n)!}$$

Takin g into account that $\;\binom m1=m\;$ , we get:

$$\require{cancel}\frac{a_{k+1}}{a_k}=\frac{\binom n{k+1}(k+1)!\cancel m\frac{(n+m-(k+1)-1)!}{(m+n)!}}{\binom nkk!\cancel m\frac{(n+m-k-1)!}{(m+n)!}}=\frac{\frac{\cancel{n!}}{\cancel{(k+1)!}(n-k-1)!}\cancel{(k+1)!}\frac{(n+m-k-2)!}{\cancel{(m+n)!}}}{\frac{\cancel{n!}}{\cancel {k!}(n-k)!}\cancel{k!}\frac{(n+m-k-1)!}{\cancel{(m+n)!}}}=$$$${}$$

$$=\frac{\frac{(n+m-k-2)!}{(n-k-1)!}}{\frac{(n+m-k-1)!}{(n-k)!}}=\frac{n-k}{n+m-k-1}=\frac{n-k}{n-k+m-1}<1\;\text{, because}\;\;m-1\ge1$$

and thus $\;a_{k+1}<a_k\;$ , and no need of calculus at all...which would be very hard to apply anyway as the above is a function of a non-cotninuous variable

• It's unbelievable I did not think of this. I spent 15 minutes doing subtraction rather than division! – астон вілла олоф мэллбэрг Sep 14 '16 at 6:31
• Wow. Where can I learn this? Can you please point some sources. I had no idea that decreasing functions can be solved like this. – roang Sep 14 '16 at 6:36