From 4 white and 5 black balls, in how many ways can 2 black balls be selected?

Correct answer: 10

The condition requires both selected balls to be black, so only the five black balls participate in the count. Because the order of selection does not matter, this is a combinations problem rather than a permutations problem. The number of choices is C(5,2) = 5!/(2!3!) = (5 × 4)/2 = 10. Therefore option A is correct. The four white balls are irrelevant because no white ball is allowed in the selected pair. A common error is to use P(5,2) = 5 × 4 = 20, which treats the same pair as different when its order is reversed; that produces option B. Values 25 and 40 do not arise from either the appropriate combination formula or the restricted selection described in the question.