TITLE: Solve an equation which contains binomial coefficient
QUESTION [5 upvotes]: I have the following equation: 
$${{x}\choose{3}}=10$$
I want to solve it for $x$ so I wrote: 
$${{x}\choose{3}}=\frac{x!}{3!(x-3)!}=10$$
which follows: 
$$\frac{x!}{(x-3)!}=60$$
I wonder how to proceed from here and solve for $x$? As $(x-3)!\neq x!-3!$.

REPLY [4 votes]: $$x!=1\cdot2\cdot3\cdot(...)\cdot(x-3)\cdot(x-2)\cdot(x-1)\cdot x$$
$$3!(x-3)!=6\cdot1\cdot2\cdot3\cdot(...)\cdot(x-3)$$
Hence
$$\frac{x!}{3!(x-3)!}=\frac{x(x-2)(x-1)}{6}=10$$
and solve from there.