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The number of ways 1st, 2nd, and 3rd place prizes could be a

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GMAT Club Legend
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Joined: 07 Jun 2014
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GRE 1: Q167 V156
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The number of ways 1st, 2nd, and 3rd place prizes could be a [#permalink] New post 05 Aug 2018, 14:51
Expert's post
00:00

Question Stats:

64% (00:24) correct 35% (00:12) wrong based on 14 sessions
Quantity A
Quantity B
The number of ways 1st, 2nd, and 3rd place prizes could be awarded to 3 out of 6 contestants
The number of ways 1st, 2nd, 3rd, 4th, and 5th place prizes could be awarded to 5 contestants


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: The number of ways 1st, 2nd, and 3rd place prizes could be a [#permalink] New post 11 Aug 2018, 13:45
Here the order of the awarded contestants matter, so we use the permutation formula:

N perm K = N!/(N-K)!

6 perm 3 = 6!/(6-3)! = 6!/3! = 6*5*4

and

5 perm 5 = 5!/(5-5)! = 5!/0! = 5*4*3*2 = 5*4*6

So they are equal. C.
GMAT Club Legend
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Joined: 07 Jun 2014
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GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Kudos [?]: 1613 [0], given: 375

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Re: The number of ways 1st, 2nd, and 3rd place prizes could be a [#permalink] New post 23 Aug 2018, 07:40
Expert's post
Explanation

In this problem, order matters; if Mari comes in 1st place and Rohit comes in 2nd, there is a different outcome than when Rohit places 1st and Mari places 2nd. Use the fundamental counting principle to solve. To determine Quantity A, make three slots (one for each prize). Six people are available to win 1st, and then five people could win 2nd, and four people could win 3rd:

654


Multiply: (6)(5)(4) = 120.

For Quantity B, make 5 slots, one for each prize. Five people can win 1st prize, then 4 people can win 2nd prize, and so on:

54321


Multiply (5)(4)(3)(2)(1) = 120. The two quantities are equal.
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Re: The number of ways 1st, 2nd, and 3rd place prizes could be a   [#permalink] 23 Aug 2018, 07:40
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