There are four contestants (John, Mark, Sally, Betty) competing for six different prizes. Each contestant will win exactly one prize. In how many different ways can the contestants be awarded the prizes?

A. 15 B. 30 C. 60 D. 180 E. 360

(E) This is a PERMUTATION question. A permutation is an ordered sequence of objects taken from a set of distinct objects without replacement. This problem is a permutation problem, and not a combination problem, because it matters which contestant the prizes go to. In a combination problem, we would merely be choosing 4 prizes out of the 6 to award (without assigning them to specific people).

In this case, if there are 6 possible prizes for the first contestant, then there are 5 possible prizes left for the second contestant, 4 prizes left for the third contestant, and 3 left for the fourth. To find the total number of ways the prizes can be awarded, we need to multiply these numbers:6 × 5 × 4 × 3 = 360.The correct answer is choice (E).

Alternative Method (Permutation Formula):The formula determines the total number of permutations given n total objects (total prizes) and k objects (number of prizes awarded) to be put into a sequence. In this question, n = 6 and k = 4.The formula is: Number of Permutations = n!/(n – k)! Number of Permutations: 6! / (6 – 4)! = 6! / 2! = (6 × 5 × 4 × 3 × 2 × 1) / (2 × 1)= 6 × 5 × 4 × 3 = 360. Again, we see that the correct answer is choice (E).

The permutation formula in the book (pg152) is different than the explanation, n! / k!(n!-k!). When do I know to omit the k! in the divisor? i.e.: (6! × 5! × 4! × 3! × 2! × 1!) / ((4! × 3! × 2! × 1!) × (2! × 1!)) = 15

The formula you use is not a formula for permutations but a formula for combinations. It is important to understand a difference between those.

Let’s say we choose a pair of two numbers (a, b) out of n numbers. If pair (1, 4) differs from pair (4, 1) then we use permutation formula. If pairs (1, 4) and (4, 1) are equivalent then we use combination formula.

In this case if John, Mark, Sally and Betty took first, third, fourth and sixth prizes, then the situation: John – first, Mark – third, Sally – fourth, Betty – sixth differs from situation: John – third, Mark – first, Sally – sixth, Betty – fourth.

Therefore we use permutation formula. If such situations were equivalent we would use combination formula instead.

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