Find all the positive integers a, b, and c such for which $\binom{a}{b} \binom{b}{c} = 2\binom{a}{c}$ I tried an the following equivalence from a different users post from a while back
that stated the following.
$\binom{a}{b} \binom{b}{c} = \binom{a}{c} \binom{a-c}{b-c}$
where does this equivalence come from?
After applying this equivalence and trying to hammer it out algebraically 
I end up with...
$\frac{(a-c)!}{(b-c)!(a-b)!}=2$
Not any closer than when I didn't use the equivalence.
How can I solve this?

You've reduced the equation to finding solutions to
$$\binom{a-c}{b-c} = 2. $$
Since $a,b,c$ are positive integers, and the only binomial coefficient equal to $2$ is $\binom{2}{1}$, the equalities
$$ a - c = 2$$ and $$b-c = 1$$ are forced.
Thus the solution set consists of consecutive triples of positive integers
$$(a,b,c) = (n+2,n+1,n) \qquad (\text{for }n\ge 0).$$
You can substitute this back into your original equation and verify that this solution set is valid.
