# Find $r$ which maximizes $\binom{20}r\binom{20}0+\binom{20}{r-1}\binom{20}1+\binom{20}{r-2}\binom{20}2+\cdots+\binom{20}0\binom{20}r$

The value of $$r$$ for which $$\binom{20}r\binom{20}0+\binom{20}{r-1}\binom{20}1+\binom{20}{r-2}\binom{20}2+\cdots+\binom{20}0\binom{20}r$$ is maximum is?

I tried to wrap my head around the solution but I don't get it. Could someone help me with it in an easier way? The solution arbitrary begins by considering the expansion of $$(1+x)^{20}$$ and then multiplying it to itself. A more question-oriented solution would be appreciated. Thanks.

$$r$$ has to be some sort of plain integer. This whole question isn't about negative or fractional indices.

• You can get a vertically centred and properly spaced ellipsis using \cdots. – joriki Apr 9 at 13:02
• @joriki What's that supposed to mean? – Rew Apr 9 at 13:53
• It supposes to mean that your post will look better if you replace .... with \cdots. – user Apr 9 at 14:17
• It means: The ellipsis that you simulated with four periods is aligned on the baseline, it isn't vertically centred with the $+$ signs and there's no space between it and the $+$ signs. That's because periods are not meant to be used like that in $\TeX$ / MathJax. With \cdots, you get a vertically centred ellipsis, and it's automatically correctly spaced because it's recognized as an operand between the $+$ signs. – joriki Apr 9 at 14:28
• @user gotcha thanks. Looks better now – Rew Apr 9 at 15:54

This expression counts the ways to select $$r$$ from $$40$$ elements by splitting the $$40$$ elements into two groups of $$20$$ each and summing over all ways to divide up $$r$$ between the two groups. Thus this is $$\binom{40}r$$.

• Woah you have given me a new vision! Thanks! – Rew Apr 9 at 13:15

The coefficient of $$x^m$$ in $$(1+x^a)^n(x^b+1)^n$$

is $$\displaystyle\sum_{r=0}^m\binom nr\binom n{m-r} x^{(a-b)r+bm}$$

Let us set $$a-b=0, b=1$$

So, we need the coefficient of $$x^m$$ in

$$(1+x)^n(x+1)^n=(1+x)^{2n}$$ which is $$\binom{2n}m$$

Here $$n=20$$

• Something goes wrong here if the last expression should mean the value of the sum in question. – user Apr 9 at 13:40
• @user, Thanks for your feedback. Please vlidate – lab bhattacharjee Apr 9 at 14:08
• I see now why your final expression is in error: you have computed the value of $\sum_m\binom nm\binom n{m-r}$, whereas the question was about $\sum_m\binom nm\binom n{\color{red}{r-m}}$. – user Apr 10 at 12:33
• @user, Please find the updated post. – lab bhattacharjee Apr 11 at 12:51
• Now it looks perfect, though for better understanding of a reader I would not change the meaning of the parameter $r$ (which is the value of interest). – user Apr 11 at 22:48