From (6) different sweets, (3) sweets are to be selected and one special sweet must be included. How many ways are there?

Correct answer: (10)

The direct answer is option B, 10. The special sweet must be included, so we do not choose it again. It is already fixed as one member of the three-sweet selection. We need two more sweets, and there are 5 ordinary sweets available. Therefore the number of choices is C2 = 5!/(2!3!) = (5×4)/2 = 10. Option A, 6, does not count all possible pairs of ordinary sweets. Option B, 10, is correct because each pair added to the special sweet gives one valid selection. Option C, 15, is the total number of ways to choose any 2 sweets from 6, so it ignores the condition that the special sweet must be included. Option D, 20, overcounts and does not follow the required fixed-special-item method. The important idea is that a compulsory item is selected first, leaving fewer items and fewer places to fill. Memory cue: “must include” means fix that item, then choose the remaining items.