Find all positive integers $a$, $b$, and $c$ for which $a \choose b$ $b \choose c$ = 2$a \choose c$

Question

Find all positive integers $a$, $b$, and $c$ for which $a \choose b$ $b \choose c$ = 2$a \choose c$.

Using the theorem ${n! \over k!(n-k)!} = {n \choose k}$ I simplified this down to $(a-c)! = 2(a-b)!(b-c)!$ but can't tell if this is close to being solved. I did write a short program to calculate all of the possible answers for every $a,b,c$ up to 10 and found that $a=b+1$ and $a=c+2$, which I believe is the answer but I can't prove it yet. Any help?

Answer 1

HINT: $\dbinom{a}b\dbinom{b}c=\dbinom{a}c\dbinom{a-c}{b-c}$; you can verify this either algebraically or with a combinatorial argument.