The students at Natural High School sell coupon books to raise money

METHOD-1

Since there are 10 homeroom and we need to award to only two of the 10 homeroom so the homerooms to be awarded can be selected in 10C2 ways = 45 ways

Now the two homerooms have to be placed at position First and Position second which can happen in = 2! ways

From the first homeroom three awardee out of 8 Students can be selected in 8C3 ways = 56 ways

The three different awards of $50, $30 and $20 can be given to the selected 3 awardee students in = 3! = 6 ways

From the Second homeroom three awardee out of 8 Students can be selected in 8C3 ways = 56 ways

But there is no need to arrange the awards among them as all the awards are same $10 = 1 way to arrange

Total Ways to make this even of award giving happen = (10C2 *2!) * (8C3 *3!) * (8C3)*1)
Total Ways to make this even of award giving happen = (45 *2* 56*6) * (56*1) = (9*5 *2* 8*7*3*2) * (8*7) = \(2^8*3^3*7^2*5\)

Answer: Option B

METHOD-2

The two homerooms out of 10 for position 1 and 2 can be identified in = 10*9 ways
Three Awardees out of 8 of First positioned homeroom can be placed in = 8*7*6 ways [All awards are different]
Three Awardees out of 8 of Second positioned homeroom can be placed in = (8*7*6)/3! ways [Divided by 3! because All awards are same $10 so no arrangement required]

Total ways to give awards = 10*9*8*7*6*[8*7*6/3!] = \(2^8*3^3*7^2*5\)

Answer: Option B

Total number of selections from 10 homerooms = 10C2 and they can be further selected in 2C1 ways to have either 3 $10 prizes or 1 each of 50,30 and 20$ prizes . Thus total ways (1) = 10C2*2C1

Now, after selecting the rooms above, number of ways of awarding 3 different (50,30,20$ prizes) out of 8 students (2) = 8C3 * 3! (factor of 3! to account for 3 different prizes, this could have been 8P3 as well)

Finally, selecting 3 students for 3 10$ prizes out of 8 students (3) = 8C3

Thus, total arrangements possible = 1 X 2 X 3 = 10C2*2C1*8C3*3!*8C3 = \((2^8 )(3^3 )(5)(7^2 )\).

B is the correct answer
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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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