Question

The value of $\left(2.{}^1{P}_0-3.{}^2{P}_1+4.{}^3{P}_2-.......upto{51}^{st}term\right)+\left(1!-2!+3!-.......upto{51}^{st}term\right)$ is equal to

• A. $1-51\left(51!\right)$
• B. $1+52!$
• C. $1$
• D. $1+51!$

Answer 1

The correct option is B

$1+52!$

Explanation for the correct answer:

The formula for ${}^n{P}_r=\frac{n!}{\left(n-r\right)!}$

$\Rightarrow$ ${}^n{P}_{n-1}=\frac{n!}{(n-n+1)!}=n!$

$\Rightarrow$ $2.{}^1{P}_0=2\times 1!=2!$

$\Rightarrow$ $3.{}^2{P}_1=3\times 2!=3!$

$\Rightarrow$ $4.{}^{3}P_2=4\times 3!=4!$

$\Rightarrow$ $52.{}^{51}P_{50}=52\times 51!=52!$

$\Rightarrow$ $S=\left(2.{}^1{P}_0-3.{}^2{P}_1+4.{}^3{P}_2-.......+52.{}^{51}{P}_{50}\right)+\left(1!-2!+3!-........+51!\right)$

$\Rightarrow$ $S=\left(2!-3!+4!-.....+52!\right)+\left(1!-2!+3!-........+51!\right)$

$\Rightarrow$ $S=1+52!$

Hence, the value of the given expression is $1+52!$.

Hence, option $\left(B\right)$ is correct .