NIMCET 2024 — Mathematics PYQ

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Question

NIMCET 2024 — Mathematics PYQ

NIMCET | Mathematics | 2024

The value of $\sum_{r = 1}^{n} \frac{1}{2 ^{n}} \cdot \frac{n P _{r}}{r !}$ is:

Choose the correct answer:

Explanation

We use the relation between permutations and combinations:

\frac{^nP_r}{r!} = \binom{n}{r}

Substituting this into the summation:

S = \sum_{r=1}^n \frac{1}{2^n} \binom{n}{r}

Since $\frac{1}{2^n}$ is a constant with respect to $r$ , it can be moved outside the summation:

S = \frac{1}{2^n} \sum_{r=1}^n \binom{n}{r}

Using the identity for the sum of binomial coefficients:

\sum_{r=0}^n \binom{n}{r} = 2^n

\binom{n}{0} + \sum_{r=1}^n \binom{n}{r} = 2^n

\sum_{r=1}^n \binom{n}{r} = 2^n - 1

(Since $\binom{n}{0} = 1$ )

Substituting this value back into the expression for $S$ :

S = \frac{2^n - 1}{2^n}

S = \frac{2^n}{2^n} - \frac{1}{2^n}

S = 1 - \frac{1}{2^n}

Using the property of exponents $\frac{1}{a^m} = a^{-m}$ :

S = 1 - 2^{-n}

Correct Option: (B)