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

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Manager
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If Bob and Jen are two of 5 participants in a race, how many [#permalink] New post 23 Oct 2006, 09:43
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?
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 [#permalink] New post 23 Oct 2006, 09:46
I got 33 here

24+6+2+1, is it rite? :oops:
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Re: Race [#permalink] New post 23 Oct 2006, 09:56
girikorat 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?


It should be 4! = 24. Basically since Jen and Bob's positions are fixed - we can just treat them as ONE combination. Taking the remaining three folks + ONE combination of Jen and Bob you have 4 permutations for the race to finish.
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 [#permalink] New post 23 Oct 2006, 10:15
Is it 60?
When Jen is in posion 1:
4! ways to arrange the other 4.

When Jim is in pos 2:
Bob has to be in 3rd, 4th or 5th pos. And the other 3 can be positioned in 3! ways
So 3x3!

When Jen is in pos 3:
Bob has to be in 4th ot 5th. Others can be positioned in 3! ways
so 2x3!

When Jen is in 4th:
Bob has to the 5th person. Others- 3!

So total = 4! + 3x3! + 2x3! + 3!
= 4! + 3!x6 = 24 + 36 = 60
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 [#permalink] New post 23 Oct 2006, 10:20
yeah its 60....I think I got it now....

its 4*3!+3*3!+2*3!+1*3! = 60


Thanks guys 8-)
  [#permalink] 23 Oct 2006, 10:20
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If Bob and Jen are two of 5 participants in a race, how many

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