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Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each

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Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each [#permalink]

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New post 19 Jul 2017, 21:32
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Difficulty:

  95% (hard)

Question Stats:

44% (01:19) correct 56% (02:13) wrong based on 97 sessions

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Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each of them finishes the marathon and no two or three athletes finish at the same time, in how many different possible orders can the athletes finish the marathon so that Alice finishes before Bobby and Bobby before Cindy?

A) 18
B) 20
C) 24
D) 30
E) 36

Source - Math Revolution

Please kindly explain your workings
[Reveal] Spoiler: OA

Kudos [?]: 61 [0], given: 276

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Re: Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each [#permalink]

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New post 20 Jul 2017, 01:58
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Without any constrains there a 5! combinations,

The constraints removes 2 degrees of freedom so 3!

the answer is 5!/3!= 5*4= 20

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Re: Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each [#permalink]

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New post 20 Jul 2017, 02:07
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ssr300 wrote:
Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each of them finishes the marathon and no two or three athletes finish at the same time, in how many different possible orders can the athletes finish the marathon so that Alice finishes before Bobby and Bobby before Cindy?

A) 18
B) 20
C) 24
D) 30
E) 36

Source - Math Revolution

Please kindly explain your workings


All 5 can be arranged in 5! Ways but without any conditions..
But A,B and C can be arranged in 3! Amongst themselves and ONLY one out of the 3! Will have ABC in that order amongst themselves.
So divide all by 3! \(\frac{5!}{3!}\)=20
B
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each [#permalink]

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New post 22 Jul 2017, 23:33
chetan2u wrote:
ssr300 wrote:
Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each of them finishes the marathon and no two or three athletes finish at the same time, in how many different possible orders can the athletes finish the marathon so that Alice finishes before Bobby and Bobby before Cindy?

A) 18
B) 20
C) 24
D) 30
E) 36

Source - Math Revolution

Please kindly explain your workings


All 5 can be arranged in 5! Ways but without any conditions..
But A,B and C can be arranged in 3! Amongst themselves and ONLY one out of the 3! Will have ABC in that order amongst themselves.
So divide all by 3! \(\frac{5!}{3!}\)=20
B


hi chetan,

why do we need to divide by 3! ?
can you explain in more detail please.

thanks!

Kudos [?]: 14 [0], given: 80

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Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each [#permalink]

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New post 23 Jul 2017, 00:37
GMATAspirer09 wrote:
chetan2u wrote:
ssr300 wrote:
Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each of them finishes the marathon and no two or three athletes finish at the same time, in how many different possible orders can the athletes finish the marathon so that Alice finishes before Bobby and Bobby before Cindy?

A) 18
B) 20
C) 24
D) 30
E) 36

Source - Math Revolution

Please kindly explain your workings


All 5 can be arranged in 5! Ways but without any conditions..
But A,B and C can be arranged in 3! Amongst themselves and ONLY one out of the 3! Will have ABC in that order amongst themselves.
So divide all by 3! \(\frac{5!}{3!}\)=20
B


hi chetan,

why do we need to divide by 3! ?
can you explain in more detail please.

thanks!




We need to divide by 3! in order to ensure that ABC order is maintained.
If we don't divide by 3 ! then ABC will be arranged in 3! ways = BAC, CAB, CBA , etc .AND we don't want this. We just want ABC order. so In order to ensure that arrangement be only in ABC , we need to divide it by 3 !

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Re: Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each [#permalink]

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New post 23 Jul 2017, 01:54
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Shouldn't the answer be just 6?

Considering "ABC" as one set, where have 3 things to arrange"ABC","D" and "E", which can be arranged in 6 ways

Here are the 6 sets:

D ABC E
E ABC D
D E ABC
E D ABC
ABC D E
ABC E D

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Re: Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each [#permalink]

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New post 01 Nov 2017, 08:05
Novice90 wrote:
Shouldn't the answer be just 6?

Considering "ABC" as one set, where have 3 things to arrange"ABC","D" and "E", which can be arranged in 6 ways

Here are the 6 sets:

D ABC E
E ABC D
D E ABC
E D ABC
ABC D E
ABC E D


Question stem says , A should be before B and B should be before C. It doesnot says they all come in line.

So possible combinations:

A B C D E
A B C E D
A D B C E
A E B C D
A B D E C
A B E D C
..... and so on.

Total will be 20 possible orders.

Answer: B

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Joined: 01 Feb 2017
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Kudos [?]: 19 [0], given: 20

Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each [#permalink]

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New post 01 Nov 2017, 15:30
Order for slotting A,B,C is fixed i.e.1.

But Order for D&E is not under any constraint and hence is 5P2.

(Position of A-B-C will be moved, retaining their constrained order, to accommodate D&E in any slot from 5P2).

Therefore, Total Possible Orders of arrangements are=
1x5P2 = 20.

Hence, Ans B

Posted from my mobile device

Kudos [?]: 19 [0], given: 20

Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each   [#permalink] 01 Nov 2017, 15:30
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