From 5 white and 7 black balls, 4 balls are to be selected and all must be black. How many selections are possible?

Correct answer: 35

The phrase “all must be black” restricts the eligible objects before counting begins. The five white balls cannot appear in any valid selection, so only the seven black balls matter. We must choose four of those seven balls, and their order does not matter because the result is a selection rather than an arrangement. Therefore the required number is C(7,4) = 7!/(4!3!) = (7×6×5)/(3×2×1) = 35. Thus option B is correct. Choosing C(12,4) would ignore the colour restriction and count invalid groups containing white balls. Using a permutation would also be unsuitable, since rearranging the same four selected balls does not create a different selection.