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.

$\endgroup$
5
  • 1
    $\begingroup$ You can get a vertically centred and properly spaced ellipsis using \cdots. $\endgroup$
    – joriki
    Apr 9, 2020 at 13:02
  • $\begingroup$ @joriki What's that supposed to mean? $\endgroup$
    – Rew
    Apr 9, 2020 at 13:53
  • 1
    $\begingroup$ It supposes to mean that your post will look better if you replace .... with \cdots. $\endgroup$
    – user
    Apr 9, 2020 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, 2020 at 14:28
  • $\begingroup$ @user gotcha thanks. Looks better now $\endgroup$
    – Rew
    Apr 9, 2020 at 15:54

2 Answers 2

Reset to default
10
$\begingroup$

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 $\binom{40}r$.

$\endgroup$
1
  • $\begingroup$ Woah you have given me a new vision! Thanks! $\endgroup$
    – Rew
    Apr 9, 2020 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 $n=20$

$\endgroup$
5
  • $\begingroup$ Something goes wrong here if the last expression should mean the value of the sum in question. $\endgroup$
    – user
    Apr 9, 2020 at 13:40
  • $\begingroup$ @user, Thanks for your feedback. Please vlidate $\endgroup$
    – lab bhattacharjee
    Apr 9, 2020 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, 2020 at 12:33
  • $\begingroup$ @user, Please find the updated post. $\endgroup$
    – lab bhattacharjee
    Apr 11, 2020 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, 2020 at 22:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .