Question

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

• A. (10, 9)
• B. (10, 7)
• C. (9, 10)
• D. (5, 10)

Answer 1

Answer: (C)

$\frac{2}{9!}+\frac{2}{3!7!}+\frac{1}{5!5!}=\frac{2^a}{b!}$
$\Rightarrow \frac{1}{7!}\left[\frac{2}{9\times 8}+\frac{2}{6}+\frac{7\times 6}{5!}\right]=\frac{2^a}{b!}$
$\Rightarrow \frac{1}{7!}\left[\frac{128}{180}\right]=\frac{1}{7!}\left[\frac{2^7}{9\times 20}\right]$
$=\frac{2^7}{7!\times 9\times 10\times 2}=\frac{2^6}{10\times 9\times 7!}$
$=\frac{2^6\times 2^3}{10\times 9\times 7!\times 2^3}$
$=\frac{2^9}{10\times 9\times 8\times 7!}=\frac{2^9}{10!}=\frac{2^a}{b!}$
Hence order pair $(a,b)=(9,10)$