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If $\frac{2}{9 !}+\frac{2}{3 ! 7 !}+\frac{1}{5 ! 5 !}=\frac{2^{a}}{b !}$, where $a, b \in N$, then the ordered pair $(a, b)$ is
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Verified Answer
The correct answer is:
$(9,10)$
We have,
$$
\begin{aligned}
& \frac{2}{9 !}+\frac{2}{3 ! 7 !}+\frac{1}{5 ! 5 !}=\frac{2^{a}}{b !} \\
\Rightarrow & \frac{1}{9 !}\left[2+\frac{2 \cdot 8 \cdot 9}{3 !}+\frac{6 \cdot 7 \cdot 8 \cdot 9}{5 !}\right]=\frac{2^{a}}{b !} \\
\Rightarrow & \frac{1}{9 !}\left[2+24+\frac{126}{5}\right]=\frac{2^{a}}{b !} \\
\Rightarrow & \frac{1}{9 !}\left[26+\frac{126}{5}\right]=\frac{2^{a}}{b !} \\
\Rightarrow & \frac{1}{9 !}\left[\frac{256}{5}\right]=\frac{2^{a}}{b !} \\
\Rightarrow & \frac{1}{9 !} \frac{256 \times 2}{5 \times 2}=\frac{512}{10 !}=\frac{2^{9}}{10 !}=\frac{2^{a}}{b !} \Rightarrow a=9 \text { and } b=10
\end{aligned}
$$
$$
\begin{aligned}
& \frac{2}{9 !}+\frac{2}{3 ! 7 !}+\frac{1}{5 ! 5 !}=\frac{2^{a}}{b !} \\
\Rightarrow & \frac{1}{9 !}\left[2+\frac{2 \cdot 8 \cdot 9}{3 !}+\frac{6 \cdot 7 \cdot 8 \cdot 9}{5 !}\right]=\frac{2^{a}}{b !} \\
\Rightarrow & \frac{1}{9 !}\left[2+24+\frac{126}{5}\right]=\frac{2^{a}}{b !} \\
\Rightarrow & \frac{1}{9 !}\left[26+\frac{126}{5}\right]=\frac{2^{a}}{b !} \\
\Rightarrow & \frac{1}{9 !}\left[\frac{256}{5}\right]=\frac{2^{a}}{b !} \\
\Rightarrow & \frac{1}{9 !} \frac{256 \times 2}{5 \times 2}=\frac{512}{10 !}=\frac{2^{9}}{10 !}=\frac{2^{a}}{b !} \Rightarrow a=9 \text { and } b=10
\end{aligned}
$$
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