Question
Prove thart.
$\frac{1}{9!}+\frac{1}{10!}+\frac{1}{11!}=\frac{122}{11!}$
$\frac{1}{9!}+\frac{1}{10!}+\frac{1}{11!}=\frac{122}{11!}$
$=\frac{1}{9!}+\frac{1}{10\times9!}+\frac{1}{11\times10\times9!}$
$=\frac{11\times10+11+1}{11\times10\times9!}$
$=\frac{110+11+1}{11!}$
$=\frac{122}{11!}$
$\text{R.H.S.}$
Hence, $\frac{1}{9!}+\frac{1}{10!}+\frac{1}{11!}= \frac{122}{11!}$
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Two events A and B which are not mutually exclusive.