Prove combinatoric inequality: ${n \choose {j+k}}\le {n \choose j}{{n-j}\choose k}$ How can one prove the following combinatoric inequality?
$${n \choose {j+k}}\le {n \choose j}{{n-j}\choose k}$$
My line of thought was:
$n$ people applied for an interview for a company. (And the company must accept everyone for some job or another)
On the right hand side, we first filtered the not so good people for simple jobs, and then sent the better ones for better jobs.
But on the left side, we just randomly chose a job for each applicant.
Now how do I conclude from here, or is the analogy incorrect?
Absolutely different solutions are welcome as well.
 A: Hint: try showing $$\binom{n}{j}\binom{n-j}{k}=\binom{n}{j+k}\binom{j+k}{j}.$$
A: A combinatorial argument something like the one that you tried is possible. 
You have $n$ applicants for two jobs, and you want to choose $j$ of them for one job and $k$ of them for the other. No one is to get two jobs. There are $\binom{n}j$ ways to pick the $j$ applicants for the first job, and once that’s been done there are $\binom{n-j}k$ ways to pick the $k$ applicants for the second job, so there are $\binom{n}j\binom{n-j}k$ ways to fill the two jobs. Alternatively, we could pick $j+k$ of the applicants to fill the two jobs, and then decide which of them get the first job and which get the second. There are $\binom{n}{j+k}$ ways to pick $j+k$ of them. No matter what $j$ and $k$ are, there’s at least one way to split the single group of $j+k$ people into a group of $j$ for the first job and a group of $k$ for the second job, so
$$\binom{n}{j+k}\le\binom{n}j\binom{n-j}k\;.$$
(The factor of $\binom{j+k}j$ in Rebecca’s solution is of course the number of ways to split the chosen $j+k$ people into the desired two groups.)
