Bunuel wrote:

The students at Natural High School sell coupon books to raise money for after-school programs. At the end of the coupon sale, the school selects six students to win prizes as follows:

From the homeroom with the highest total coupon-book sales, the students with the first-, second- and third-highest sales receive $50, $30, and $20, respectively; from the homeroom with the second-highest total coupon-book sales, the three highest-selling students receive $10 each. If Natural High School has ten different homerooms with eight students each, in how many different ways could the six prizes be awarded? (Assume that there are no ties, either among students or among homerooms.)

A. \((2^7)(3^2)(5)(7^2)\)

B. \((2^8)(3^3)(5)(7^2)\)

C. \((2^9)(3)(5^2)(7^2)\)

D. \((2^9)(3^4)(5)(7^2)\)

E. \((2^8)(3^5)(5)(7^2)\)

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION: There are three separate sets of decisions to be made in this problem. You can think of these decisions as questions to answer. First, which two homerooms have the highest total sales? Second, who are the three highest-selling students from the first-place homeroom? Third, who are the three highestselling students from the second-place homeroom? In each of these decisions, the crucial issue is whether order matters. In other words, does switching the order of the choices have any effect on the result? If switching the order matters, the choice is that of a "permutation." If switching does not matter, the choice is that of a "combination."

First, choose the two homerooms. Here, order matters, because the first-place homeroom receives different prizes than does the second-place homeroom. The slot method (fundamental counting principle) is simplest: the two homerooms can be chosen in 10*9 different ways. (Do not bother computing the product, even though it is easy to do, because we are going to factor down to primes anyway.) The anagram method works too, using anagrams of the "word" 12NNNNNNNN: 10!/(8!) = 10*9.

Second, select the three prize winners from the first-place homeroom. Here, order also matters, because the three selected students receive three different prizes. The slot method (fundamental counting principle) is simplest again: 8*7*6 different ways. Alternatively, use the anagram method with the "word" 123NNNNN: 8!/(5!) = 8*7*6.

Finally, select the three prize winners from the second-place homeroom. In this case, order does not matter, because the same prize is given to each of the three winning students. Therefore, this is a combination, using the anagram method for the "word" YYYNNNNN: 8!/(5!3!) = 8*7 different ways.

Since these three decisions are sequential, the total number of ways in which the winners can be chosen is:

\((10*9)*(8*7*6)*(8*7) = 2^8*3^3*5*7^2\)

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