Requesting deeper understanding of binomial coefficient

Question

I noticed that $\binom {52} 4$ * $\binom {48} 1$ is $5$ times that of $\binom {52} 5$. So for example, if we were to draw $4$ cards from a standard deck then draw $1$ more, we cannot just say there are $\binom {52} 4$ * $\binom {48} 1$ ways to do that because we counted $5$ times too high that way. So my question is what is a more general formula for this type of thing?

Suppose "k" is anywhere from $1$ to $4$ cards on the first "group" so we have $\binom {52} k$, so then the 2nd group would be $\binom {52-k} {5-k}$, but what would the "correction factor" be in the semi-general case? For example, in the $\binom {52} 4$ * $\binom {48} 1$ case it was $5$x too high so we need to divide by $5$ to get it to equal $\binom {52} 5$. What about if we only chose $3$ cards the first go round or only $2$...?

Next suppose we didn't restrict ourselves to having $52$ cards to choose from but rather n cards and we choose k of them such that k is no more than half of n and n is at least $10$. Then what can we say generically about "busting" up our draw into $2$ "groups" and then "correcting" them?

Answer 1

Suppose that I have a set $S$ of $n$ white balls. The calculation

$$\binom{n}k\binom{n-k}\ell$$

counts the ways to perform the following operation: pick $k$ of the balls in $S$ and paint them red, then pick $\ell$ of the remaining white balls and color them blue. I could get the same results by first picking any $k+\ell$ balls, then choosing $k$ of those to paint red while painting the other $\ell$ blue. The ways to perform this operation are counted by the expression

$$\binom{n}{k+\ell}\binom{k+\ell}k\;.$$

Thus,

$$\binom{n}k\binom{n-k}\ell=\binom{n}{k+\ell}\binom{k+\ell}k\;,$$

and therefore

$$\binom{n}{k+\ell}=\frac{\binom{n}k\binom{n-k}\ell}{\binom{k+\ell}k}\;.$$

In your original problem $n=52$, $k=4$, and $\ell=1$.