4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ You can get a vertically centred and properly spaced ellipsis using \cdots. $\endgroup$ – joriki Apr 9 at 13:02
  • $\begingroup$ @joriki What's that supposed to mean? $\endgroup$ – Rew Apr 9 at 13:53
  • 1
    $\begingroup$ It supposes to mean that your post will look better if you replace .... with \cdots. $\endgroup$ – user Apr 9 at 14:17
  • 1
    $\begingroup$ 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. $\endgroup$ – joriki Apr 9 at 14:28
  • $\begingroup$ @user gotcha thanks. Looks better now $\endgroup$ – Rew Apr 9 at 15:54
9
$\begingroup$

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 $\binom{40}r$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Woah you have given me a new vision! Thanks! $\endgroup$ – Rew Apr 9 at 13:15
4
$\begingroup$

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 $n=20$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Something goes wrong here if the last expression should mean the value of the sum in question. $\endgroup$ – user Apr 9 at 13:40
  • $\begingroup$ @user, Thanks for your feedback. Please vlidate $\endgroup$ – lab bhattacharjee Apr 9 at 14:08
  • $\begingroup$ 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 $\sum_m\binom nm\binom n{\color{red}{r-m}}$. $\endgroup$ – user Apr 10 at 12:33
  • $\begingroup$ @user, Please find the updated post. $\endgroup$ – lab bhattacharjee Apr 11 at 12:51
  • $\begingroup$ 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). $\endgroup$ – user Apr 11 at 22:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.