In how many different ways can the top eight new indie bands : GMAT Problem Solving (PS)
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# In how many different ways can the top eight new indie bands

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In how many different ways can the top eight new indie bands [#permalink]

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16 Jun 2010, 10:03
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In how many different ways can the top eight new indie bands be ranked on a top eight list?

The top hit song for each of the eight bands will compete to receive monetary awards of $1000,$500, $250, and$100; respectively. In how many ways can the awards be given out?
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16 Jun 2010, 10:21
dlwilson1122 wrote:
In how many different ways can the top eight new indie bands be ranked on a top eight list? the top hit song for each of the eight bands will compete to receive monetary awards of $1000,$500, $250, and$100; respectively. In how many ways can the awards be given out?

would you multiply the prizes to determine how many ways?

this is confusing to me - thanks for any help / explanation

thanks

Welcome to Gmat Club! Below is the solution for your problem

In how many different ways can the top eight new indie bands be ranked on a top eight list?

# of permutations of $$n$$ distinct objects is $$n!$$, so the answer is $$8!$$.

The top hit song for each of the eight bands will compete to receive monetary awards of $1000,$500, $250, and$100; respectively. In how many ways can the awards be given out?

# of ways to choose $$n$$ objects from $$k$$ distinct opbjects when order matters is $$P^n_k=\frac{k!}{(k-n)!}$$.

So for our original question as there will be 4 awards, then # of ways to choose 4 bands out of 8, when the order matters is $$P^4_8=5*6*7*8$$.

For more on this issue please check Combinatorics chapter of Math Book (link in my signature).

Hope it helps.
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16 Jun 2010, 10:25
thanks for explaining
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16 Jun 2010, 10:26
Using the multiplication principle there will be 8! ways.
i.e for the first place there are 8 possible ways, for the second place only 7 possible ways will remain and so forth till the 8th place there will be 1 way. So in total there will be 8! = 8x7x6x5x4x3x2x1 ways.

For the second part of the question, order matters and hence this is a permutation problem.
There will be 8P4 ways
8P4 = 8!/4! = 8x7x6x4 ways.
Re: Probability   [#permalink] 16 Jun 2010, 10:26
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