Question

If $\left(\frac{2}{9!}\right)+$ $\left(\frac{2}{3!7!}\right)$ $+\frac{1}{5!5!}=\frac{2^a}{b!}$, where $a,b\in N$, then the ordered pair $(a,b)$ is

• A. $(9,10)$
• B. $(10,9)$
• C. $(7,10)$
• D. $(10,7)$

Answer 1

The correct option is A

$(9,10)$

Explanation for the correct options:

Finding the value:

Given Data:

$\left(\frac{2}{9!}\right)+$ $\left(\frac{2}{3!7!}\right)$ $+\frac{1}{5!5!}=\frac{2^a}{b!}$

$\Rightarrow$$\frac{1}{7!}$$[\frac{2}{9\times 8}+\frac{2}{6}+\frac{7\times 6}{5!}]=\frac{2^a}{b!}$

$\Rightarrow$ $\frac{1}{7!}\times \frac{128}{180}=\frac{2^a}{b!}$

$\Rightarrow$ $\frac{2^7}{7!\times 9\times 10\times 2}=\frac{2^a}{b!}$

Multiply numerator and denominator by $2^3$

$\Rightarrow$ $\frac{2^6\times 2^3}{10\times 9\times 2^3\times 7!}$$=\frac{2^a}{b!}$

$\Rightarrow$ $\frac{2^9}{10!}=\frac{2^a}{b!}$

So, $a=9$and $b=10$

The ordered pair $(a,b)$ is $(9,10).$

Hence Option (A) is correct.