Combinatorial Probability

Combinatorial probability counts favourable selections over total selections using combinations, appropriate when order does not matter in the choice. This selection-without-order pattern recurs throughout JEE Advanced. The total number of ways to choose 3 people from 12 is C(12, 3) = 220. Favourable selections require exactly two men from seven and one woman from five, giving C(7, 2) × C(5, 1) = 21 × 5 = 105. Hence the probability is 105/220, which simplifies by dividing numerator and denominator by 5 to 21/44. Option 7/22 = 70/220 undercounts by using C(7,2) = 7. Option 35/44 wildly overstates the favourable fraction. Option 5/22 = 50/220 mismatches the men-and-women split. The method applies the product rule across independent sub-selections, since choosing the men and choosing the women are separate stages, and the equally-likely-combinations definition of probability over the C(12,3) outcomes. It is worth noting that order is irrelevant here, which is exactly why combinations rather than permutations are used; had the three seats been distinct roles, the counting would instead involve arrangements. Plausibility check: 21/44 ≈ 0.477 is a legitimate probability strictly below 1, and since two-men-one-woman is just one of several possible compositions, a value near one-half is reasonable given that men slightly outnumber women in the committee and this split is among the more balanced selections.