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JEE MATHEMATICS MCQ: COMBINATIONS

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If 29!+23!7!+15!5!=2ab!,\frac{2}{{9!}} + \frac{2}{{3!7!}} + \frac{1}{{5!5!}} = \frac{{{2^a}}}{{b!}}, where a,b,∈Na,b, \in N, then the ordered pair (a,b)\left( {a,b} \right) is
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Correct option: A
(9,10)\left( {9,10} \right)
Option A ((9,10)) is correct because simplifying the given expression results in 2910!\frac{{{2^9}}}{{10!}}.

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