Question

Find the value of the following expression: \[\dfrac{1}{9!}+\dfrac{1}{3!7!}+\dfrac{1}{5!5!}+\dfrac{1}{7!3!}+\dfrac{1}{9!}\]
A. $\dfrac{{{2}^{9}}}{10!}$
B. $\dfrac{{{2}^{10}}}{8!}$
C. $\dfrac{{{2}^{11}}}{9!}$
D. $\dfrac{{{2}^{10}}}{7!}$
E. $\dfrac{{{2}^{8}}}{9!}$

Answer 1

Hint:
 We will first start with bringing the common terms together and then we will add them by taking the LCM of the denominator and then convert the numerator such that the common terms get canceled. Then we will write the numerator in the form of ${{2}^{n}}$ and write the denominator in the form of factorial.

Complete step by step answer:
We know that $n!=1\times 2\times 3\times ............\times n$
And we are given in the question that: \[\dfrac{1}{9!}+\dfrac{1}{3!7!}+\dfrac{1}{5!5!}+\dfrac{1}{7!3!}+\dfrac{1}{9!}\]
Now let’s arrange the given expression and write it as: \[\left( \dfrac{1}{9!}+\dfrac{1}{9!} \right)+\left( \dfrac{1}{3!7!}+\dfrac{1}{7!3!} \right)+\dfrac{1}{5!5!}\] ,
After adding the terms in the brackets we will get: \[\left( \dfrac{2}{9!} \right)+\left( \dfrac{2}{3!7!} \right)+\dfrac{1}{5!5!}\]
Now we will solve this by adding these fractions:
\[\dfrac{\left( 2\times 3!7!5!5! \right)+\left( 2\times 9!5!5! \right)+\left( 9!3!7! \right)}{9!3!7!5!5!}\text{ }.............\text{ Equation 1}\text{.}\]
Now we will write $9!5!5!=7!\times \left( 8\times 9 \right)\times 5!\times 5!=7!\times \left( 4\times 2\times 3\times 3 \right)\times 5!\times 5!=7!\times \left( 4\times 6\times 3 \right)\times 5!\times 5!$ , now we know that the value of $3!=3\times 2\times 1=6$ , replacing $6$as $3!$,
Therefore:
$\begin{align}
  & 7!\times \left( 4\times 6\times 3 \right)\times 5!\times 5!=7!\times \left( 4\times 3!\times 3 \right)\times 5!\times 5!=12\times 7!\times 3!\times 5!\times 5! \\
 & \Rightarrow \left( 9!\times 5!\times 5! \right)=12\times 7!\times 3!\times 5!\times 5!\text{ }............\text{ Equation 2}\text{.} \\
\end{align}$
Now we will write \[9!3!7!=\left( 5!\times 6\times 7\times 8\times 9 \right)\times 3!\times 7!=3024\times 5!\times 3!\times 7!\] ,
 Therefore: \[9!3!7!=3024\times 5!\times 3!\times 7!\text{ }.............\text{ Equation 3}\text{.}\]
Now, we will put the values from equation 2 and 3 into equation 1:
$\begin{align}
  & \dfrac{\left( 2\times 3!7!5! \right)+\left( 2\times 9!5!5! \right)+\left( 9!3!7! \right)}{9!3!7!5!5!} \\
 & =\dfrac{\left( 2\times 3!7!5! \right)+\left( 2\times 12\times 7!\times 3!\times 5!\times 5! \right)+\left( 3024\times 5!\times 3!\times 7! \right)}{9!3!7!5!5!} \\
\end{align}$
Now, we will take out $3!\times 7!\times 5!$ common from the above fraction:
$\begin{align}
  & \dfrac{\left( 2\times 3!7!5!5! \right)+\left( 2\times 12\times 7!\times 3!\times 5!\times 5! \right)+\left( 3024\times 5!\times 3!\times 7! \right)}{9!3!7!5!5!} \\
 & =\left( \dfrac{2\times 3!7!5!5!}{9!3!7!5!5!} \right)+\left( \dfrac{24\times 3!7!5!5!}{9!3!7!5!5!} \right)+\left( \dfrac{3024\times 3!7!5!}{9!3!7!5!5!} \right) \\
\end{align}$

Now we will cancel $3!\times 7!\times 5!$ from the numerator and the denominator:
$\begin{align}
  & \Rightarrow \left( \dfrac{2\times 3!7!5!5!}{9!3!7!5!5!} \right)+\left( \dfrac{24\times 3!7!5!5!}{9!3!7!5!5!} \right)+\left( \dfrac{3024\times 3!7!5!}{9!3!7!5!5!} \right)=\dfrac{2}{9!}+\dfrac{24}{9!}+\dfrac{3024}{9!5!} \\
 & \Rightarrow \dfrac{2}{9!}+\dfrac{24}{9!}+\dfrac{3024}{9!5!}=\dfrac{26}{9!}+\dfrac{3024}{9!5!}=\dfrac{\left( 26\times 5! \right)+3024}{9!5!} \\
 & \Rightarrow \dfrac{6144}{9!5!} \\
\end{align}$
Now we will expand $5!=5\times 4\times 3\times 2\times 1=10\times 12$ and replace it in the above fraction:
\[\Rightarrow \dfrac{6144}{9!5!}=\dfrac{6144}{9!\times 10\times 12}\]
We will replace $9!\times 10=10!$, therefore: \[\dfrac{6144}{9!\times 10\times 12}=\dfrac{512}{10!}\] , Now we will write $512$ as${{2}^{9}}$
And therefore: \[\dfrac{512}{10!}=\dfrac{{{2}^{9}}}{10!}\] .
Therefore, the value of \[\dfrac{1}{9!}+\dfrac{1}{3!7!}+\dfrac{1}{5!5!}+\dfrac{1}{7!3!}+\dfrac{1}{9!}=\dfrac{{{2}^{9}}}{10!}\]
Hence, the correct option is A.

Note:
Note that, $n!=\left( n-1 \right)!\times n$, students can make mistakes while expanding all the factorials because the result obtained will be a very big number and can create difficulty in the calculations. So always try and take out the common factor while converting the given factorial.