Question

If $\frac{1}{1!11!}+\frac{1}{3!9!}+\frac{1}{5!7!}=\frac{2^n}{m!}$ , then the value of $m+n$ is

• A. $18$
• B. $23$
• C. $12$
• D. $22$

Answer 1

Answer: (D)

$\frac{1}{12!}\left\{\frac{12!}{1!11!}+\frac{12!}{3!9!}+\frac{12!}{5!7!}\right\}=\frac{2^n}{m!}$
$\Rightarrow \frac{1}{12!}\left\{{}^{12}{C}_1{+}^{12}{C}_3{+}^{12}{C}_5\right\}=\frac{2^n}{m!}$
$\Rightarrow \frac{1}{12!}\left\{{}^{12}{C}_1{+}^{12}{C}_1{+}^{12}{C}_3{+}^{12}{C}_3{+}^{12}{C}_5{+}^{12}{C}_5\right\}=\frac{2^{n+1}}{m!}$
$\Rightarrow \frac{1}{12!}\left\{{}^{12}{C}_1{+}^{12}{C}_{11}{+}^{12}{C}_3{+}^{12}{C}_9{+}^{12}{C}_5{+}^{12}{C}_7\right\}=\frac{2^{n+1}}{m!}$
$\Rightarrow \frac{1}{12!}\left\{{}^{12}{C}_1{+}^{12}{C}_3{+}^{12}{C}_5{+}^{12}{C}_7{+}^{12}{C}_9{+}^{12}{C}_{11}\right\}=\frac{2^{n+1}}{m!}$
$\Rightarrow \frac{2^{11}}{12!}=\frac{2^{n+1}}{m!}\Rightarrow n=10,m=12$
$\Rightarrow m+n=22$