Tough and lengthy question:

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Tough and lengthy question:

by \'manpreet singh » Mon Jan 07, 2013 11:17 pm
Q.1)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.) Write your answer as a
product of primes raised to various powers (do not actually compute the number)?

-First of all i would like to state,when i read the question i could not grasp everything ,so i had to re-read.Wasting some time, still it was not fully clear.I am clearly struggling with lengthy questions ,in which the question tries to bore you down.Need little help regarding this and yes, this particular problem also seemed tough to me when i got it finally.Do explain the solution....

-If some one can explain me the usage of anagram method and where to apply it, i will be grateful, i think its a good tool,if only i could master it.
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by e-GMAT » Tue Jan 08, 2013 2:54 am
'manpreet singh wrote:Q.1)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.) Write your answer as a
product of primes raised to various powers (do not actually compute the number)?

-First of all i would like to state,when i read the question i could not grasp everything ,so i had to re-read.Wasting some time, still it was not fully clear.I am clearly struggling with lengthy questions ,in which the question tries to bore you down.Need little help regarding this and yes, this particular problem also seemed tough to me when i got it finally.Do explain the solution....

-If some one can explain me the usage of anagram method and where to apply it, i will be grateful, i think its a good tool,if only i could master it.
Hi,

As per my understanding...

1. There are 10 HomeRooms in the schools
2. Each HomeRoom has 8 students
3. There are 6 students, who will be awarded
-----a. 3 students out of 8 students from any of the 10 HomeRooms. This HomeRooms will be selected among a competition of 10 HomeRooms. There will be I, II, & III prizes for the students of that HomeRoom.
-----b. 3 students out of 8 students from any of the remaining 9 HomeRooms ( II Best HomeRoom). Note that there will not be any I, II, & III prize for students, since all 3 prizes are of $10 each.

=> # of ways of choosing I Homeroom (The Winner)
# of ways = 10C1
# of ways of choosing 3 winners = 8P3 ( Order matters here)
So, # of ways of choosing 3 winners from I Homeroom = 10C1*8P3

=> # of ways of choosing II Homeroom (II Best)
# of ways = 9C1 ( Now only 9 HomeRooms are remaining)
# of ways of choosing 3 winners = 8C3 ( Order does not matter here. All prizes are eual)
So, # of ways of choosing 3 winners from II Homeroom = 9C1*8C3

=> Total # of ways of choosing 6 winners = 10C1*8P3*9C1*8C3 = 2^8*3^3*5*7^2

Hope it helps.

-Shalabh Jain

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by \'manpreet singh » Thu Jan 10, 2013 9:52 pm
Thanks for easy explanation.

It definetely helps Shalabh, though from above solution i have one more thing to clarify that when ever order is important we take permutation to find the total no of ways of choosing,where as anywhere order is not important we take combination.Correct me if my conclusion is wrong here!

Br,
Manpreet