Reference: Appendix C.1-11, CLRS 4th edition.
Argue that for any integers $n,j,k \geq 0$ and $j+k \leq n$ $$\binom{n}{j+k} \leq \binom{n}{j}\binom{n-j}{k}$$.
I can see why this inequality holds from algebraic proof clearly. From combinatorial proof, I've converted statement into this funny way,
Suppose a boy is given option to choose $j+k$ chocolate from $n$ chocolates and on RHS a boy is asked to choose first $j$ chocolates from $n$ and then $k$ chocolates from remaining $n-j$ chocolates.
Question: At the end boy will choose $j+k$ chocolates then why number of ways of choosing $j+k$ chocolates together are less than choosing them in parts (first $j$ and then $k$ from remaining $n-j$)?
I have a brief argument to it when boy is choosing $j$ then $k$ he'll have to to look up or consider options again in $n-j$ chocolates which he may have seen in first $n \choose j$ expression. I don't know whether it makes any sense or not.
For $\binom {n}{k+j}$, you end up with a set of $k+j$ chocolates which are all indistinguishable from one another.
For $\binom{n}{j}\times \binom{n-j}{k}$, you end up with an ordered pair ($S,T$), where $S$ is a set of $j$ chocolates, and $T$ is a set of $k$ chocolates which is disjoint from $S$. This means that the chocolates in $S$ are distinguishable from the chocolates in $T$.
Since method $2$ specifies a little more information about the chocolates than method $1$, there are more ways to perform this method $2$.
For example, in the case where $n=8,j=2,k=3$, one way to complete the first method would be to choose five numbers: $$ \{2,4,5,6,8\} $$ However, there are perform method $2$ which correspond to that some subset. Namely, you could do $$ (\{2,4\},\{5,6,8\}),\quad (\{2,5\},\{4,6,8\}),\quad (\{2,6\},\{4,5,8\}),\quad \dots $$ This shows that for each output of method $1$, there are several outputs of method $2$, which shows that there are more ways to to method $2$.