# Finding maximum value of ${n \choose r}$ for given value of n [duplicate]

While I was solving some binomial theorem chapter questions I encountered many questions which asked me me to find maximum value of ${n \choose r}$ for given value of n.

Example: Find n for which $$n!(21-n)!$$ has minimum value ?

I didn't get how to do this without substituting so I looked at the solution , it was given

" $n!(21-n)! =21!/{21 \choose n}$

Thus, for minimum value of $n!(21-n)!$ , ${21 \choose n}$ should be maximum, which is for n=10. "

So I wanted to ask ,

Is there a general result to get the value of r for which ${n \choose r}$ is maximum for given value of n ?

• I am confused. Of course it is clear which value of $r$ gives the maximum. So what is your difficulty? If you are unclear, find the corresponding $r$ for $n=2,3,4,5,6$ and see the pattern. – almagest Jul 3 '16 at 11:31
• So are you actually asking what is the maximum value for $f(n, m)$, given $m$, when $f(n, m) = n!\left(m - n!\right)$? – Jared Jul 3 '16 at 11:34
• Do I need to substitute every time the values of r for finding maximum value of ${n \choose r}$ for given n? – Dimenein Jul 3 '16 at 11:36
• en.wikipedia.org/wiki/Pascal%27s_triangle If $n$ is even, $r=n/2$. If $n$ is odd, $r=(n\pm1)/2$. Example: if $n=21$, ${n\choose r}$ is maximal at $r=10$ and $r=11$. – Did Jul 3 '16 at 11:39
• Actually my function is $f(x,y) ={x \choose y}$ ,and my question is find maximum value of $f(x,y)$ for given $x$. – Dimenein Jul 3 '16 at 11:39

For $n$ even, $r=n/2$. For $n$ odd, the maximum occurs at $r=(n-1)/2$ and at $r=(n+1)/2$.

Proof: Let's exclude trivial cases assuming that $n\ge 3$. Since $\binom nk=\binom n{n-k}$ it suffices to show that $\binom nk<\binom nr$ for $k<r$, where $r=\lfloor n/2 \rfloor$.

But $$\binom nk=\frac{n(n-1)\cdots(n-k+1)}{k!}$$ $$\binom nr=\frac{n(n-1)\cdots(n-r+1)}{r!}$$ Dividing, $$\frac{\binom nr}{\binom nk}=\frac{(n-k)(n-k-1)\cdots(n-r+1)}{r(r-1)\cdots(k+1)}$$

Note that $$n-r+1\ge n-\frac{n+1}2+1=\frac{n+1}2> r$$ Since the last factor of the numerator is greater than the first factor of the denominator, we have that every factor of the numerator is greater than every factor of the denominator, so this quotient is clearly $>1$.

Assume that $l+k=n$ and consider $x=l!\times k!$. We are to find the values of $l$ and $k$ such that $x$ gets its minimum value.

Dividing $$x=l!\times k!=1\times 2\times \cdots\times l\ \ \ \color{red}{\times}\ \ \ 1\times 2\times \cdots \times k$$ by

$$y=(l-1)!\times (k+1)!=1\times 2\times \cdots\times (l-1) \ \ \ \color{red}{\times}\ \ \ 1\times 2\times \cdots \times (k+1)$$

we have $\frac{x}{y}=\frac{l}{k+1}$ as the result. It is seen that $x>y$ if and only if $l>k+1$. So, the minimum of $x$ is obtained when $l$ and $k$ are the closest possible values to each other.

In your case, we have $l+k=21$, and so the minimum is obtained when $l=10$ and $k=11$.