Meg and Bob are among the 5 participants in a cycling race. : GMAT Problem Solving (PS)
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# Meg and Bob are among the 5 participants in a cycling race.

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Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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22 Oct 2007, 19:14
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Difficulty:

35% (medium)

Question Stats:

65% (01:00) correct 35% (01:23) wrong based on 188 sessions

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Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob?

A. 24
B. 30
C. 60
D. 90
E. 120

OPEN DISCUSSION OF THIS QUESTION IS HERE: meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Jul 2014, 08:28, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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22 Oct 2007, 19:22
IrinaOK wrote:
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

JXXXX => Bob can be in any X...Total 4! = 24
XJXXX => Bob can be in the last 3 Xs...Total 3! = 6
XXJXX => 2! = 2
XXXJX => 1! = 1

24+6+2+1 = 33
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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22 Oct 2007, 19:31
bkk145 wrote:
IrinaOK wrote:
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

JXXXX => Bob can be in any X...Total 4! = 24
XJXXX => Bob can be in the last 3 Xs...Total 3! = 6
XXJXX => 2! = 2
XXXJX => 1! = 1

24+6+2+1 = 33

Yep I missed factorials for arranagements
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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22 Oct 2007, 20:34
Let me redeem myself:

JXXXX => Bob can be in any X...Total 4! = 24

XJXXX => Bob can be in the last 3 Xs...Total 3! = 6. However, since the first person can change too and it cannot be Bob, it can only be 3 other people. So Total = 6*3 = 18

XXJBX => 3! = 6 total
XXJXB => 3! = 6 total
So this yields total of 12

XXXJB => 3! = 6

Ans = 24+18+12+6 = 60
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16 Nov 2007, 22:24
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Refer to one method promoted by Johnr, i figured out a quick solution to this problem:

with 5 participants, there're 5! ways to arrange them.

Because Bob and Jen have identical roles in these 5 participants, the chance that each of them finishes before the another is 50%. Thus, the number of ways in which Jen finishes before Bob is 5! * 50% = 60 ways
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16 Nov 2007, 22:31
IrinaOK wrote:
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

J has to finish before B (given)

J finishes 1st then B can finish in 4 ways and other three can finish in 3! ways to give us 4*3! ways = 24
J finishes 2nd then B can finish in 3 ways and other three can finish in 3! ways to give us 3*3! ways = 18
J finishes 3rd then B can finish in 2 ways and other three can finish in 3! ways to give us 2*3! ways = 12
J finishes 4th then B can finish in 1 way (last) and other three can finish in 3! ways to give us 1*3! ways = 6

Total = 24 + 18 + 12+ 6 = 60
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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17 Nov 2007, 16:07
What's the assessment of level of difficulty of this prob? Range one would see this problem in terms of GMAT scoring?
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17 Nov 2007, 16:42
IrinaOK wrote:
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

for me, the question is ambigious in the above red part, basically "infront of Bob". does it mean "JB" or "JXXXB"?

if "infront of Bob" means only "JB", then there are 24 (4x3!) ways.
if "infront of Bob" means only "JXXXB", then there are 60 [3!(1+2+3+4)]ways.
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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07 Sep 2009, 11:34
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There is another way to get this solved. This method can be generalized to solve such kind of problems.

The steps are:
1. Take a specific case that satisfies the given condition.
2. Find the possible arrangements for that case.
3. Find the number of such cases.

Let me elaborate with the solution.

1. We can choose any two positions out of the 5 positions. Say i choose a specific case where Jen is 1st & Bob is 4th.
J _ _ B _
2. The three blank spaces can be filled in 3! ways.
3. Now 2 positions can be selected from 5 positions ( as done in step 1 ) in 5C2 ways

So, total ways: 3! * 5C2 = 60
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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28 Sep 2009, 10:25
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

Soln: total number of ways is 5! = 120 ways

Of these , in exactly half of the arrangements, will Jen finish ahead of Bob. Hence
= 120/2
= 60 ways
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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01 Oct 2009, 21:03
We can count Bob & Jen as one contestant.

So the total number of permutations is 4!, which is 24.

Did not read the question right.
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02 May 2011, 22:30
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out of 5! = 120 ways.half of the ways are when B is ahead of J.
hence 60
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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03 May 2011, 00:20
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There are always 2 ways to arrange Jen and Bob.
First possibility: J(...)B (with or without people in between)
Second possibility: B(...)J (with or without people in between)

Solution: 5!/2 = 60
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26 Jul 2014, 05:55
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Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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26 Jul 2014, 08:29
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IrinaOK wrote:
Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob?

A. 24
B. 30
C. 60
D. 90
E. 120

Total # of ways the race can be finished is 5!. In half of the cases Meg finishes ahead of Bob and in other half Bob finishes ahead of Meg. So, ways Meg to finish ahead of Bob is 5!/2=60.

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six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html
in-how-many-different-ways-can-the-letters-a-a-b-91460.html
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OPEN DISCUSSION OF THIS QUESTION IS HERE: meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
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Meg and Bob are among the 5 participants in a cycling race.   [#permalink] 26 Jul 2014, 08:29
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