Question
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 $$\left( {a,b} \right)$$ is
A.
$$\left( {9,10} \right)$$
B.
$$\left( {10,9} \right)$$
C.
$$\left( {7,10} \right)$$
D.
$$\left( {10,7} \right)$$
Answer :
$$\left( {9,10} \right)$$
Solution :
$$\eqalign{
& \frac{2}{{9!}} + \frac{2}{{3!7!}} + \frac{1}{{5!5!}} \cr
& = \frac{1}{{1!9!}} + \frac{1}{{3!7!}} + \frac{1}{{5!5!}} + \frac{1}{{3!7!}} + \frac{1}{{9!1!}} \cr
& = \frac{1}{{10!}}\left\{ {^{10}{C_1} + {\,^{10}}{C_3} + {\,^{10}}{C_5} + {\,^{10}}{C_7} + {\,^{10}}{C_9}} \right\} \cr
& = \frac{1}{{10!}}\left( {{2^{10 - 1}}} \right) = \frac{{{2^9}}}{{10!}} = \frac{{{2^a}}}{{b!}}\left( {{\text{given}}} \right) \cr
& \Rightarrow a = 9,b = 10 \cr} $$