Prove $\binom km\binom tk=\binom tm\binom{t-m}{t-k}$

Question

$$\binom km\binom tk=\binom tm\binom{t-m}{t-k}\\m\le k\le t$$ I'm not sure where to start. Combinatorics has confused me, as there doesn't seem to be a place in the textbook to read about it, and the solutions have been far too difficult to think about. Is there a strategy to this?

Answer 1

If what you want is a scenario or a procedure that might help you understand what the identity is about, here's one:

Imagine you're a producer of a sitcom and your task is to select $\color{blue} m$ cast members out of $\color{red} t$ candidates via a two-stage process: first you narrow it down to $\color{orange} k$ candidates, then out of these $\color{orange} k$ you get the final $\color{blue} m$. The cast should have a good chemistry so you'll pick a combo of $\color{blue} m$ together all at once as opposed to picking a person for each role one by one.

How many possible ways are there to do this? It's the L.H.S. of the identity $${t \choose k}{k \choose m}$$

At the same time, the cruel reality of the TV industry is that you actually have already decided the cast (of size $\color{blue} m$ out of $\color{red} t$), and the two-stage process is just a formality. Here what you need to do (to put up a show for the public) is to pick ${\color{orange} k}-\color{blue} m$ people (who you are going to discard anyway in the 2nd stage) out of the pool of `losers' of size ${\color{red} t}-\color{blue} m$, to team them up with the predetermined $\color{blue} m$ to form the $\color{orange} k$ temporary winners of the 1st stage.

How many possible situations are there for this? It's the R.H.S. of the identity $${t \choose m}{t-m \choose k-m}$$

Answer 2

Expand the binomial coefficients into factorials: $$\binom km\binom tk=\frac{k!t!}{m!k!(k-m)!(t-k)!}$$ This simplifies to $$\frac{t!}{m!(k-m)!(t-k)!}$$ Multiply both halves of the fraction by $(t-m)!$, then rearrange: $$\frac{\color{blue}{t!}(t-m)!}{\color{blue}{m!(t-m)!}(k-m)!(t-k)!}=\color{blue}{\binom tm}\binom{t-m}{t-k}$$ Hence $$\binom km\binom tk\equiv\binom tm\binom{t-m}{t-k}$$

Answer 3

Expanding them as factorials,

\begin{eqnarray} \binom{k}{m}\binom{t}{k}=\frac{\color{red}{k!}}{m!(k-m)!}\frac{t!}{\color{red}{k!}(t-k)!} &=&\frac{t!}{m!(k-m)!(t-k)!}\\ &=&\frac{t!\color{red}{(t-m)!}}{m!\color{red}{(t-m)!}(k-m)!(t-k)!}\\ &=&\frac{\color{orange}{t!}\color{green}{(t-m)!}}{\color{orange}{m!{(t-m)!}}\color{green}{(k-m)!(t-k)!}}\\ &=&\frac{t!}{m!(t-m)!}\frac{(t-m)!}{(t-k)!(k-m)!}=\binom{t}{m}\binom{t-m}{t-k} \end{eqnarray}

---

(This is basically the same as Parcly's. I just wanted to try out the colors like he did.)