Binomial coefficient (7 choose 3)=Binomial coefficient (7 choose x) How can I solve the following equation:

$$\frac{7!}{3!(7-3)!}=\frac{7!}{x!(7-x)!}$$

What seems problematic to me in particular is the term $$x!(7-x)!$$.
you have from the denominator that 
$$x!(7-x)! = 3! 4! = 4! 3!$$
Choose $x = 4$.
You have to solve $$x!(7-x)! = 3!4!$$

Note that $5$ does not divide $3!4!$. Thus $x&lt;5$ and $7-x &lt; 5 \iff x &gt; 2$.

So $x=3$ and $x=4$ are the solutions.