If $P_{m}$ stands for $\ ^{m}P_{m}$, then $1 + 1P_{1} + 2P_{2} + 3P_{3} + ... + n.P_{n}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
$1 + 1.P_{1} + 2.P_{2} + 3.P_{3} + \ldots + n.P_{n}$
$= 1 + 1.(1!) + 2.(2!) + \ldots + n(n!)$
$= 1 + (2 - 1)1! + (3 - 1)2! + \ldots + \left( (n + 1) - 1 \right)n!$
$= 1 + 2! - 1! + 3! - 2! + ... + (n + 1)! - n!$
$= (n + 1)!$

Q. If $P_{m}$ stands for $\ ^{m}P_{m}$, then $1 + 1P_{1} + 2P_{2} + 3P_{3} + ... + n.P_{n}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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