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(3) different prizes are to be given to (6) students, and no student can receive more than one prize. How many ways are possible?

Answer and explanation

Correct answer: (120)

The direct answer is C: 120. The three prizes are different, so assigning prize 1, prize 2 and prize 3 is an ordered selection of three different students. Choose the recipient of the first prize in 6 ways, the second in 5 ways, and the third in 4 ways. The total is \(6\times5\times4={}^{6}P_3=120\). Option C is correct. Option A, 18, is too small. Option B, 60, does not count all assignments. Option D, 216, is \(6^3\) and would allow one student to receive multiple prizes, which the question forbids. Since prizes are distinct, order matters even though recipients cannot repeat.

Related tags

PermutationsPrizesOrdered Selection

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