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)

(9, 10)

Answer 2

$\frac{2}{9!} + \frac{2}{3! \,7!}+\frac{1}{5! \,5!} =\frac{2^{a}}{b!}$
$\Rightarrow \frac{1}{7!} \left[\frac{2}{9 \times8}+\frac{2}{6}+\frac{7\times 6}{5!}\right]=\frac{2^{a}}{b!}$
$\Rightarrow \frac{1}{7!}\left[\frac{128}{180}\right]=\frac{1}{7!}\left[\frac{2^{7}}{9\times 20}\right]$
$=\frac{2^{7}}{7! \times 9\times 10\times 2}=\frac{2^{6}}{10\times 9\times 7!}$
$= \frac{2^{6}\times 2^{3}}{10\times 9\times 7!\times 2^{3}}$
$=\frac{2^{9}}{10\times 9\times 8\times 7!}=\frac{2^{9}}{10!}=\frac{2^{a}}{b!}$
Hence order pair $(a, b) = (9,10)$