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  4. If (1/1 ! 11 !)+(1/3 ! 9 !)+(1/5 ! 7 !)=(2n/m !) , then the value of m+n is

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

NTA AbhyasNTA Abhyas 2020Binomial Theorem

Solution:

$\frac{1}{12 !}\left\{\frac{12 !}{1 ! 11 !} + \frac{12 !}{3 ! 9 !} + \frac{12 !}{5 ! 7 !}\right\}=\frac{2^{n}}{m !}$
$\Rightarrow \frac{1}{12 !}\left\{^{12} C_{1} + ^{12} C_{3} + ^{12} C_{5}\right\}=\frac{2^{n}}{m !}$
$\Rightarrow \frac{1}{12 !}\left\{^{12} C_{1} + ^{12} C_{1} + ^{12} C_{3} + ^{12} C_{3} + ^{12} C_{5} + ^{12} C_{5}\right\}=\frac{2^{n + 1}}{m !}$
$\Rightarrow \frac{1}{12 !}\left\{^{12} C_{1} + ^{12} C_{11} + ^{12} C_{3} + ^{12} C_{9} + ^{12} C_{5} + ^{12} C_{7}\right\}=\frac{2^{n + 1}}{m !}$
$\Rightarrow \frac{1}{12 !}\left\{^{12} C_{1} + ^{12} C_{3} + ^{12} C_{5} + ^{12} C_{7} + ^{12} C_{9} + ^{12} C_{11}\right\}=\frac{2^{n + 1}}{m !}$
$\Rightarrow \frac{2^{11}}{12 !}=\frac{2^{n + 1}}{m !}\Rightarrow n=10,m=12$
$\Rightarrow m+n=22$