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# If Bob and Jen are two of 5 participants in a race, how many

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Director
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If Bob and Jen are two of 5 participants in a race, how many [#permalink]  22 Oct 2007, 19:14
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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?

VP
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Re: probability_Bob and Jen [#permalink]  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
VP
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Re: probability_Bob and Jen [#permalink]  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
VP
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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
SVP
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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
Director
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Re: probability_Bob and Jen [#permalink]  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
Director
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What's the assessment of level of difficulty of this prob? Range one would see this problem in terms of GMAT scoring?
Director
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Re: probability_Bob and Jen [#permalink]  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.
Manager
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Re: probability_Bob and Jen [#permalink]  07 Sep 2009, 11:34
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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Manager
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Re: probability_Bob and Jen [#permalink]  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
Senior Manager
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Re: probability_Bob and Jen [#permalink]  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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VP
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Re: probability_Bob and Jen [#permalink]  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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Manager
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Re: probability_Bob and Jen [#permalink]  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
Re: probability_Bob and Jen   [#permalink] 03 May 2011, 00:20
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