Question

The sum $S = \frac{1}{9!} + \frac{1}{3!7!} + \frac{1}{5!5!} + \frac{1}{7!3!} + \frac{1}{9!}$ is equal to

• A. $\frac {2^{10}} { 8!}$
• B. $\frac{2^{9}} {10!}$
• C. $\frac{2^{7}} {10!}$
• D. $\frac{2^{6}} { 10!}$
• E. $\frac{2^{5}} { 8!}$

Answer 1

Answer: (B)

Let
$ S =\frac{1}{9 !}+\frac{1}{3 ! 7 !}+\frac{1}{5 ! 5 !}+\frac{1}{7 ! 3 !}+\frac{1}{9 !}$
$=\frac{1}{10 !}\left[\frac{10 !}{9 !}+\frac{10 !}{3 ! 7 !}+\frac{10 !}{5 ! 5 !}+\frac{10 !}{7 ! 3 !}+\frac{10 !}{9 !}\right] $
$ =\frac{1}{10 !}\left[{ }^{10} C_{1}+{ }^{10} C_{3}+{ }^{10} C_{5}+{ }^{10} C_{7}+{ }^{10} C_{9}\right] $
$= \frac{1}{10 !}\left(2^{10-1}\right) $
$ \left[\because C_{1}+C_{3}+C_{5}+\ldots=2^{n-1}\right] $
$= \frac{2^{9}}{10 !} $