NIMCET 2024 Paper

Section: Mathematics

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Question : 27 of 120

Marks: +1, -0

The value of

∑↙{r=1}↖{n} \;{n P_r}/{2^n × r!}

is:

$2^n$
$1-2^{-n}$
$2^n-1$
$2^{2 n}-1$

Solution:

We know that

\;{{ }^n P_r}/{r!}={ }^n C_r

Hence the given sum is

\;\;{1}/{2^n} ∑↙{r=1}↖{n}{ }^n C_r=\;{1}/{2^n} ({ }^n C_1+{ }^n C_2+. . .+{ }^n C_n)

\;=\;{1}/{2^n} (2^n-1) =1-\;{1}/{2^n}

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