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Find the maximum and minimum value of the function $f(x)=\binom{16-x}{2x-1}+\binom{20-3x}{4x-5}.$

ATTEMPT:-

By A.M.-G.M. inequality, $\frac{a+b}{2}\ge\sqrt{ab}$, $\quad$ for $a,b\gt 0$ with equality at $a=b.$

For minimum value, $\binom{16-x}{2x-1}=\binom{20-3x}{4x-5}.$

$\qquad$$\implies x=2.$

But when I use my graphing calculator, I get a different result.Where am i going wrong?

And how do I calculate the maximum value?

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  • $\begingroup$ If the binomial coefficients involved were "larger" (in the sense of bigger values in the upper part), I'd be tempted to consider the normal approximation to a binomial distribution as a way of getting a continuous analog to this discrete problem. But that seems to be overkill here. You have only a few $x$ values to check, and the graphing calculator to help you, so see if that gives a better answer than your arithmetic-geometric mean approach. $\endgroup$
    – hardmath
    May 1, 2016 at 10:57
  • $\begingroup$ I don't think you explained that approach (or shared the calculator output) well enough for me to comment in more detail. $\endgroup$
    – hardmath
    May 1, 2016 at 10:57
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    $\begingroup$ @yasir : When you write $\binom nm$, does it satisfy $n\ge 0,m\ge 0$ and $n-m\ge 0$ where $n,m\in\mathbb Z$? $\endgroup$
    – mathlove
    May 1, 2016 at 11:39
  • $\begingroup$ @mathlove: It's not specified in the problem. $\endgroup$
    – miyagi_do
    May 2, 2016 at 4:19
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    $\begingroup$ @hardmath: For $\binom{n}{m}$, if $n\ge 0,m\ge 0$ and $n-m\ge 0$, where $n.m\in Z$, $x \in {{2,3}}$ with minima at $2$. and maxima at $3.$ $\endgroup$
    – miyagi_do
    May 2, 2016 at 4:29

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