Question

The maximum number of permutations of $2n$ letters in which there are only $a^{\prime }s$ and $b^{\prime }s,$ taken all at a time is given by

• A. ${}^{2n}{C}_n$
• B. $\frac{2}{1}\cdot \frac{6}{2}\cdot \frac{10}{3}\cdot \cdots \frac{4n-6}{n-1}\cdot \frac{4n-2}{n}$
• C. $\frac{n+1}{1}\cdot \frac{n+2}{2}\cdot \frac{n+3}{3}\cdot \frac{n+4}{4}\cdot \cdots \frac{2n-1}{n-1}\cdot \frac{2n}{n}$
• D. $\frac{2^n[1\cdot 3\cdot 5\cdots (2n-3\left)\right(2n-1\left)\right]}{n!}$

Answer 1

Answer: (A), (B), (C), (D)

$\frac{2^n[1\cdot 3\cdot 5\cdots (2n-3\left)\right(2n-1\left)\right]}{n!}$

Let the number of $a^{\prime }s$ be $x.$
Then, number of $b^{\prime }s$ is $2n-x$
As $a^{\prime }s$ and $b^{\prime }s$ are alike, number of permutations $=\frac{\left(2n\right)!}{x!(2n-x)!}={}^{2n}{C}_x$

Now, we know that, if $n$ is even, ${}^n{C}_r$ is maximum for $r=\frac{n}{2}$
$\therefore {}^{2n}{C}_x$ is maximum when $r=\frac{2n}{2}=n$
Hence, maximum number of permutations $={}^{2n}{C}_n$
${}^{2n}{C}_n=\frac{\left(2n\right)!}{n!n!}$
$=\frac{2n\cdot (2n-1)\cdot (2n-2)\cdots (n+1)\cdot n!}{n!n!}$
$=\frac{2n\cdot (2n-1)\cdot (2n-2)\cdots (n+1)}{n!}$
$=\frac{(n+1)\cdot (n+2)\cdot (n+3)\cdots (2n-1)\cdot \left(2n\right)}{1\cdot 2\cdot 3\cdots (n-1)\cdot n}$

${}^{2n}{C}_n=\frac{2n!}{n!n!}$

$=\frac{1\cdot 2\cdot 3\cdots (2n-1)\cdot \left(2n\right)}{n!n!}$

$=\frac{(1\cdot 3\cdot 5\cdots (2n-1\left)\right)(2\cdot 4\cdot 6\cdots 2n)}{n!n!}$

$=\frac{1\cdot 3\cdot 5\cdots (2n-1)2^n(1\cdot 2\cdot 3\cdots n)}{n!n!}$
$=\frac{2^n(1\cdot 3\cdot 5\cdots (2n-3\left)\right(2n-1)}{n!}$
$=\frac{2\cdot 6\cdot 10\cdots (4n-2)}{1\cdot 2\cdot 3\cdots \left(n\right)}$
$($multiplying whole numerator by $2^n)$
$=\frac{2}{1}\cdot \frac{6}{2}\cdot \frac{10}{3}\cdot \cdots \frac{4n-6}{n-1}\cdot \frac{4n-2}{n}$