For what value of $m$ the is sum $\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}$ where ${p\choose q} = 0$, if $p<q$, a maximum

Question

For what value of $m$ the is sum

$$\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}\text{where ${p\choose q}$} = 0\text{, if $p<q$, a maximum}$$

My approach

$$\sum_i^{m} {10 \choose i}{20 \choose m - i} = {10 \choose 0}{20 \choose m} + {10 \choose 1}{20 \choose m - 1} + \dots + {10 \choose m}{20 \choose 0}$$

$$(1 +x)^{20} = {20 \choose 0} + {20 \choose 1}x + \dots + {20 \choose m-1}x^{m-1} + {20 \choose m}x^{m} + \dots + {20 \choose 20}x^{20}$$

$$(1 +x)^{10} = {10 \choose 0} + {10 \choose 1}x + \dots + {10 \choose 10}x^{10}$$

Later what to do?? Any other method or hint will be greatly welcomed.

Answer 1

Vandermonde's Identity says that $$ \sum_{i=0}^m\binom{10}{i}\binom{20}{m-i}=\binom{30}{m}\tag1 $$ The central binomial coefficient is the greatest. Therefore, the maximum of $(1)$ is when $m=15$; that is, $\binom{30}{15}$.

Answer 2

$$(1+x)^{10}= 10_{C_{0}}+10_{C_{1}} x+\cdots+10_{C_{10}} x^{10} \ldots \ldots \ldots \ldots \ldots .(1)$$ $$(1+x)^{20}= 20_{C_{20}}x^{20}+20_{C_{19}} x^{19}+\cdots+20_{C_{0}} \ldots \ldots \ldots \ldots \ldots .(2)$$ On multiplying these two... $$(1+x)^ {30}= (\cdots)x^{30}+(\cdots)x^{29}+\cdots$$ The largest coefficient would be that of $x^{15}$ and that would be equal to ... $$\sum_{i=0}^{m}\left(\begin{array}{l}10 \\ i\end{array}\right)\left(\begin{array}{l}20 \\ 15-i\end{array}\right),$$ Hope this helps!

Answer 3

Here's how I'm gonna to approach it. Hopefully, it'll give some intuition into the Vandermonde identity that you can read up on later.

Consider a jar of $r$ red marbles and $b$ blue marbles. We need to find the number of ways to pick $m$ marbles.

Well we can pick $1$ red and $m - 1$ blue marbles or $2$ red and $m - 2$ blue marbles or $3$ red and $m - 3$ and so forth...

We need to find the sum total of the number of ways to pick $i$ red and $m - i$ blue marbles, which is exactly the RHS of: $$ {r+b \choose m} = \sum_i^m {r \choose i}{b \choose m - i}$$

And the LHS is exactly as we described previously: the number of ways to choose $m$ marbles from all $r + b$ marbles. Does this make sense? We are essentially a combinatorial trick called "counting by two ways".

Now where does ${r+b \choose m}$ achieve its maximum? Well Pascal's triangle has it's maximums of every row down it's center column, so ${r+b \choose m}$ is maximized at $\left\lfloor {\frac{r+b}{2}}\right\rfloor$

Answer 4

Write the (3) expansion as $$(1+1/x)^{10}=~^{10}C_0 + ~^{10}C_1 (1/x)+ ~^{10}C_2(1/x^2)+......+ ~^{10}C_n(1/x^{10})~~~~~(3)$$ Now multiply (2) and (3) mainly collecting the terms independent of $x$ in the product $$S=\text{coefficient of $x^0$ in} (1+x)^{20}(1+1/x)^{10}$$