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JEE MATHEMATICS MCQ: PERMUTATIONS

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Two players P1P_1 and P2P_2 play a series of '2n2 n' games. Each game can result in either a win or loss for P1P_1. Total number of ways in which P1P_1 can win the series of these games, is equal to
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Correct option: A
12(22n−2nCn)\frac{1}{2}\left(2^{2 n}-{ }^{2 n} C_n\right)
Option A (12(22n−2nCn)\frac{1}{2}\left(2^{2 n}-{ }^{2 n} C_n\right)) is correct because it represents the sum of ways P1P_1 can win more than n games, which is half of the total ways minus the ways to win exactly n games.

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