TITLE: Given that $a,b,d$ are positive integers, if $(a+b)! d! = (a+d)! b!,$ then do we have $b= d?$
QUESTION [5 upvotes]: Question: Given that $a,b,c,d$ are positive integers satisfying 
  $$\binom{a+b}{a} = \binom{c+d}{c} \quad \text{and} \quad a = c,$$
  do they imply that $b = d$?

From assumption, we have 
$$\frac{(a+b)!}{a!b!} = \frac{(c+d)!}{c!d!}$$
$$\Rightarrow \frac{(a+b)!}{b!} = \frac{(a+d)!}{d!}$$
$$\Rightarrow (a+b)! d! = (a+d)! b!.$$
Can we deduce that $b=d$ from equation above?

REPLY [11 votes]: Yes.  If $b \gt d$ you can write $$\frac{(a+b)!}{(a+d)!}=\frac{b!}{d!}$$
Each side has $b-d$ terms but all the terms on the left are $a$ greater than the corresponding term on the right, so the left will be greater than the right.  $d\gt b$ works the same.