Three distinct prizes are to be awarded to seven students. No student can receive more than one prize. How many ways are possible?

Correct answer: 210

The governing concept is permutation because the three prizes are distinct. Choosing student X for the first prize and student Y for the second is different from reversing those assignments, so order of assignment matters. There are 7 choices for the first prize, then 6 remaining choices for the second, and finally 5 remaining choices for the third. Thus the number of assignments is 7 × 6 × 5 = 210, which is also 7P3 = 7!/(7−3)!. Therefore, option A is correct. Option C, 35, would count only an unordered group of three students using 7C3. Option B is not the relevant permutation value, and option D incorrectly treats the choices as independent with repetition allowed.