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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.

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Parth Gor
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    LHS: choose $j+k$ out of $n$ boys to give chocolate to. RHS: choose $j$ out of $n$ boys to give dark chocolate to, then choose $k$ from the rest to give milk chocolate to. RHS has more possibilities because there are two type of chocolates – acat3 Apr 30 '23 at 15:05
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    This answers your question: Requesting deeper understanding of binomial coefficient Your second boy does more than choosing $j+k$ chocolates. He puts $j$ of them in his right pocket and the remaining $k$'s in his left pocket. This is why there are more possibilities. – Anne Bauval Apr 30 '23 at 15:11

2 Answers2

2

  1. For $\binom {n}{k+j}$, you end up with a set of $k+j$ chocolates which are all indistinguishable from one another.

  2. 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 $2$.

N. F. Taussig
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Mike Earnest
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1

Hint

Use the fact that

$$\binom{j+k}{j}\binom{n}{j+k} = \binom{n}{j}\binom{n-j}{k}$$

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