Long Method
As there are only 5 participants we can split this into four cases
1. 3 People between them
There are 3! ways for this to happen
6 ways
2. 2 People between them
We can select the man who is gonna stand out in 3c1 ways and arrange him either infront of the group or after the group. Again you can arrange the people between them in 2 ways
so 2*3c1*2= 12 ways
3. 1 between them
We can select the one who is gonna be between them in 3c1 ways
then we can arrange these people in 3!( two other men and the group)
So the total no: of ways =18 (3!*3)
4. None between them
we can arrange the people in 4! ways ( 3 people and a group)
that is 24 ways
So the total no: of ways.. 24+18+12+6=60
Simpler Method
As there are 5 people there are 5! ways to arrange them
ie 120 ways to arrange them ...
So, in half of them one guy has to be ahead of the other and vice versa
that gives us 60 as answer
Difficult Math Question #31 - Combinations
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Source: Beat The GMAT — Problem Solving |
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800guy
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OA:
approach: first fix Jen, and then fix Bob. Then fix the remaining three.
Case 1: When Jen is in the first place.
=> Bob can be in any of the other four places. => 4.
The remaining 3 can arrange themselves in the remaining 3 places in 3! Ways.
Hence total ways = 4*3!
Case 2: When Jen is in the second place.
=> Bob can be in any of three places => 3.
The remaining can arrange themselves in 3 places in 3! Ways.
Continuing the approach.
Answer = 4*3! +3*3! +2*3! +1*3! = 10*3! = 60 ways.
approach: first fix Jen, and then fix Bob. Then fix the remaining three.
Case 1: When Jen is in the first place.
=> Bob can be in any of the other four places. => 4.
The remaining 3 can arrange themselves in the remaining 3 places in 3! Ways.
Hence total ways = 4*3!
Case 2: When Jen is in the second place.
=> Bob can be in any of three places => 3.
The remaining can arrange themselves in 3 places in 3! Ways.
Continuing the approach.
Answer = 4*3! +3*3! +2*3! +1*3! = 10*3! = 60 ways.
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limits660
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I have a question about this one
here are what i believe to be the possibilities:
1 2 3 4 5
J B X X X
J X B X X
J X X B X
J X X X B
X J B X X
X J X B X
X J X X B
X X J B X
X X J X B
X X X J B
Therefore 10 ways that Jan finished in front of Bob
How can there be 60? Im missing something fundamental, right?
here are what i believe to be the possibilities:
1 2 3 4 5
J B X X X
J X B X X
J X X B X
J X X X B
X J B X X
X J X B X
X J X X B
X X J B X
X X J X B
X X X J B
Therefore 10 ways that Jan finished in front of Bob
How can there be 60? Im missing something fundamental, right?
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ajith
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limits660 wrote:I have a question about this one
here are what i believe to be the possibilities:
1 2 3 4 5
J B X X X
J X B X X
J X X B X
J X X X B
X J B X X
X J X B X
X J X X B
X X J B X
X X J X B
X X X J B
Therefore 10 ways that Jan finished in front of Bob
How can there be 60? Im missing something fundamental, right?
The problem here is X's you used are not identical, They can be arranged in 6 ways each.
For example Let us assume that 5 members are X,Y,Z, J ,B
then we will take your first arrangement,
J B X Y Z
this is not the only one
J B X Z Y
J B Z X Y
J B Z Y X
J B Y X Z
J B Y Z X
are also there,ie you missed 5 of those...
For each of your arrangement there are 5 more are there ..
That gives you 10*5 =50 more arrangements...
50+10 is 60 ....
Hope that helps!
Last edited by ajith on Wed Oct 25, 2006 7:01 am, edited 1 time in total.
