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?
 A: 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.
A: You have $a$ people, and you want to choose $b$ kings and $c$ super kings amongst those kings. You can either choose the kings first and then the super kings from them, or you can choose the super kings first and then choose the rest of the kings from the remaining group.
A: You correctly arrived at recognizing that $\binom{a-c}{b-c}=2$
Now, the only remaining step is to recognize that the only binomial coefficient which equals two is $\binom{2}{1}$, thus $a-c=2$ and $b-c=1$, so they form a triple $(c+2,c+1,c)$
Why is this?  You can prove it by strong induction noting that $\binom{n}{0}=1, \binom{n}{n}=1$ and $\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$ which for $n\geq 3$ is by inductive hypothesis either $0+1$ or is the sum of two positive numbers, one of which is greater than or equal to $2$, and is thus greater than or equal to three.
Alternatively, you can note that $\binom{n}{1}=\binom{n}{n-1}=n$ and binomial coefficients are monotonic up until the midpoint., that is $n=\binom{n}{1}<\binom{n}{2}<\binom{n}{3}<\dots<\binom{n}{\lfloor n/2\rfloor}$, similarly $\binom{n}{\lceil n/2\rceil}>\dots>\binom{n}{n-2}>\binom{n}{n-1}=n$
