Question

If \[\dfrac{2}{{9!}} + \dfrac{2}{{3!7!}} + \dfrac{1}{{5!5!}} = \dfrac{{{2^a}}}{{b!}}\] where \[a,b \in N\] . Then, what is the value of the order pair \[\left( {a,b} \right)\]?
A. \[\left( {9,10} \right)\]
B. \[\left( {10,9} \right)\]
C. \[\left( {7,10} \right)\]
D. \[\left( {10,7} \right)\]

Answer 1

Hint: First, simplify the left-hand side by splitting the terms. Then, rewrite the terms in the form of factorial. After that, solve the left-hand side and rewrite the numerator as the exponent of 2. In the end, compare it with the right-hand side to get the value of the required ordered pair.

Formula Used: \[n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times .... \times 3 \times 2 \times 1\]

Complete step by step solution: The given equation is \[\dfrac{2}{{9!}} + \dfrac{2}{{3!7!}} + \dfrac{1}{{5!5!}} = \dfrac{{{2^a}}}{{b!}}\] where \[a,b \in N\].

Let’s simplify the left-hand side.
 \[\dfrac{1}{{9!}} + \dfrac{1}{{9!}} + \dfrac{1}{{3!7!}} + \dfrac{1}{{3!7!}} + \dfrac{1}{{5!5!}} = \dfrac{{{2^a}}}{{b!}}\]
\[ \Rightarrow \dfrac{1}{{9!}} + \dfrac{1}{{3!7!}} + \dfrac{1}{{5!5!}} + \dfrac{1}{{3!7!}} + \dfrac{1}{{9!}} = \dfrac{{{2^a}}}{{b!}}\]
Multiply and divide the left-hand side by \[10!\].
\[\dfrac{1}{{10!}}\left[ {\dfrac{{10!}}{{1!9!}} + \dfrac{{10!}}{{3!7!}} + \dfrac{{10!}}{{5!5!}} + \dfrac{{10!}}{{7!3!}} + \dfrac{{10!}}{{9!1!}}} \right] = \dfrac{{{2^a}}}{{b!}}\]
Now rewrite the left-hand side by using the factorial formula \[n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times .... \times 3 \times 2 \times 1\].
\[\dfrac{1}{{10!}}\left[ {\dfrac{{10 \times 9!}}{{1!9!}} + \dfrac{{10 \times 9 \times 8 \times 7!}}{{3!7!}} + \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5!}}{{5!5!}} + \dfrac{{10 \times 9 \times 8 \times 7!}}{{7!3!}} + \dfrac{{10 \times 9!}}{{9!1!}}} \right] = \dfrac{{{2^a}}}{{b!}}\]
\[ \Rightarrow \dfrac{1}{{10!}}\left[ {10 + 120 + 252 + 120 + 10} \right] = \dfrac{{{2^a}}}{{b!}}\]
\[ \Rightarrow \dfrac{1}{{10!}}\left[ {512} \right] = \dfrac{{{2^a}}}{{b!}}\]
\[ \Rightarrow \dfrac{1}{{10!}}\left[ {{2^9}} \right] = \dfrac{{{2^a}}}{{b!}}\]
Now compare both sides.
We get,
\[a = 9\] and \[b = 10\]

Option ‘A’ is correct
Note: The factorial of a number is a product of all whole numbers less than that number up to 1.
Formulas:
\[n! = n\left( {n - 1} \right)\left( {n - 2} \right)....3 \times 2 \times 1\]
\[n! = n\left( {n - 1} \right)!\]