Question:

If $\frac{2}{9!} + \frac{2}{3! \,7!}+\frac{1}{5! \,5!} =\frac{2^{a}}{b!}$ where $a,b \in \, N$ then theordered pair $(a, b)$ is

Updated On: May 11, 2024
  • (10, 9)
  • (10, 7)
  • (9, 10)
  • (5, 10)
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The Correct Option is C

Solution and Explanation

$\frac{2}{9!} + \frac{2}{3! \,7!}+\frac{1}{5! \,5!} =\frac{2^{a}}{b!}$
$ \Rightarrow \frac{1}{7!} \left[\frac{2}{9 \times8}+\frac{2}{6}+\frac{7\times 6}{5!}\right]=\frac{2^{a}}{b!} $
$\Rightarrow \frac{1}{7!}\left[\frac{128}{180}\right]=\frac{1}{7!}\left[\frac{2^{7}}{9\times 20}\right]$
$ =\frac{2^{7}}{7! \times 9\times 10\times 2}=\frac{2^{6}}{10\times 9\times 7!}$
$ = \frac{2^{6}\times 2^{3}}{10\times 9\times 7!\times 2^{3}} $
$=\frac{2^{9}}{10\times 9\times 8\times 7!}=\frac{2^{9}}{10!}=\frac{2^{a}}{b!}$
Hence order pair $(a, b) = (9,10)$
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Concepts Used:

Permutations and Combinations

Permutation:

Permutation is the method or the act of arranging members of a set into an order or a sequence. 

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Combination:

Combination is the method of forming subsets by selecting data from a larger set in a way that the selection order does not matter.

  • Combination refers to the combination of about n things taken k at a time without any repetition.
  • The combination is used for a group of data where the order of data does not matter.