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JEE MATHEMATICS MCQ: PERMUTATIONS

1 Easy
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Application +4 | -1
If x∈Nx \in {N} and (x+2)!(2x−1)!⋅(2x+1)!(x+3)!=727\frac{(x+2) !}{(2 x-1) !} \cdot \frac{(2 x+1) !}{(x+3) !}=\frac{72}{7}, then xx is equal to
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Correct option: B
4
Option B (4) is correct because simplifying the given factorial equation leads to a quadratic equation, which has x=4 as a valid solution.

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