Question

$2^n{P}_n$ is .equal to

• A. $(n+1)(n+2).......\left(2n\right)$
• B. $2^n[\mathrm{1.3.5}.....(2n-1\left)\right]$
• C. $\left(2\right).\left(6\right).\left(10\right)....(4n-2)$
• D. $n!\left(2^n{C}_n\right)$

Answer 1

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

The correct options are
A $n!\left(2^n{C}_n\right)$
B $(n+1)(n+2).......\left(2n\right)$
C $2^n[\mathrm{1.3.5}.....(2n-1\left)\right]$
D $\left(2\right).\left(6\right).\left(10\right)....(4n-2)$

$2^n{P}_n=\frac{2n!}{(2n-n)!}$

as ${}^n{P}_r=\frac{n!}{n-n!}$

$=\frac{\left(2n\right)!}{n!}$

We can write it as,

$n^n\left[\mathrm{1.3.5.......}\right(2n-1\left)\right]$

or

$\mathrm{2.6.10........}(4n-2)$

also we $2^n{P}_n=\left(2n\right)(2n-2)........(n+1)$

and checking from option (d)

$n!\left(2{}^n{C}_n\right)=\frac{n!\left(2n\right)!}{n!n!}$

$=2^n{P}_n$

$\therefore$ $\left(a\right),\left(b\right),\left(c\right),\left(d\right)$ all are correct.

Hence, all the answers are correct.