Question

If ${\mathrm{P}}_{\mathrm{m}}$ stands for ${}^{\mathrm{m}}{\mathrm{P}}_{\mathrm{m}}$, then $1+1{\mathrm{P}}_1+2{\mathrm{P}}_2+3{\mathrm{P}}_3+\dots +{\mathrm{nP}}_{\mathrm{n}}$ is equal to:

• A. $1!$
• B. $(n+3)!$
• C. $(n+2)!$
• D. $(n+1)!$

Answer 1

The correct option is D

$(n+1)!$

Explanation for correct option:

${\mathrm{P}}_{\mathrm{m}}={}^{\mathrm{m}}\mathrm{P}_{\mathrm{m}=}\mathrm{m}!$

${\mathrm{p}}_1=1!,{\mathrm{p}}_2=2!,{\mathrm{p}}_3=3!,\dots ..........p_n=\mathrm{n}!$

Now,

$1+1{\mathrm{P}}_1+2{\mathrm{P}}_2+3{\mathrm{P}}_3+\dots +{\mathrm{nP}}_{\mathrm{n}}$

$=1+\left(1\right)1!+\left(2\right)2!+\left(3\right)3!+........................\dots +\left(\mathrm{n}\right)\mathrm{n}!$

$=1+(2!–1!)+(3!–2!)+(4!–3!)+........................\dots +\left(\right(n+1)!–n!)$

$=(1–1!)+(2!–2!)+(3!–3!)+(4!–4!)+\dots \dots \dots \dots \dots \dots \dots +(n!–n!)+(n+1)!$

$=(n+1)!$

Hence, Option(4) is correct