First slide

Permutations

Question

The number of positive integers $n$ such that $2^{n}$ di vides n! is

Moderate

A

exactly 1
B

exactly 2
C

infinite
D

none of these

Solution

The exponent of 2 in $n!$ is given by
$E = [\frac{n}{2}] + [\frac{n}{2^{2}}] + [\frac{n}{2^{3}}] + \dots$
where [x] denotes greatest integer $\leq x, As [x] \leq x \forall x$
$[\frac{n}{2^{m}}] = 0$ after finite number of terms. Thus we get
$E < \frac{n}{2} + \frac{n}{2^{2}} + \frac{n}{2^{3}} + \dots = \frac{n / 2}{1 - 1 / 2} = n$
Thus, there is no positive integer for which $2^{n}$ divides $n!$

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