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Challenge Question - Counting Methods

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Challenge Question - Counting Methods [#permalink] New post 19 Aug 2004, 00:43
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Hey guys

This was the first draft of a challenge question we wanted to include in challenge 2. I wrote it. :oops: I and bb could not quite agree on a solution, so this question was not included. Actual question text would have been much shorter. I am not including answer choices here..just to see how you guys approach the problem. Let me know if there are any inconsistencies.

For the summer, a group of eight friends made plans for a road trip from Houston, Texas to Washington D.C. Some of the friends wanted to stop over by their friends place on the way, so they decided to start out on two different routes. They studied various candidate routes and chose 2 routes, Route 1 and Route 2. A group of 4 will start out on Route 1 and a group of 4 will start out on Route 2. Paul and Jessica want to visit their friend in Knoxville and they can only go on Route 2. If only 4 people can be seated in one car and all the friends will take turns in driving during the trip, in how many ways can the group of eight be seated in two cars?
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 [#permalink] New post 19 Aug 2004, 01:29
The confusion I can see is when you decide the combinatino for the car travelling along route2. We know paul and jessica must take route 2, so that leaves 2 places remaining, but when we go to calculating the combinations, do we use 6C2, or 6C4 ? That's the confusing part. I'm stil trying to reason out.
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 [#permalink] New post 19 Aug 2004, 07:12
I think it's a fair question, though definately hard. But I don't think in the end it takes too much work.

If we look at the 2 cars, and start with route 1, there are 6 people who can be in the car (because Paul and Jessica aren't in it). So the number of ways way can possibly put those people in the car is 6x5x4x3.

Now four people are in the first car. There are 4 people left over, including Paul and Jessica. So the second car has 4x3x2x1 ways it can be loaded.

So the answer should be (I think) 6x5x4x3x4x3x2x1. It's not a combinations question at all, because you said everyone can drive, so each way is a different arrangement of how the people could be in the car.

Or, perhaps you mean it to be a combinations problem, and that's why everyone takes turns driving? In that case, it would be (6x5x4x3x4x3x2x1)/4!(4!) = 15
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 [#permalink] New post 19 Aug 2004, 12:22
ian7777 wrote:
I think it's a fair question, though definately hard. But I don't think in the end it takes too much work.

If we look at the 2 cars, and start with route 1, there are 6 people who can be in the car (because Paul and Jessica aren't in it). So the number of ways way can possibly put those people in the car is 6x5x4x3.

Now four people are in the first car. There are 4 people left over, including Paul and Jessica. So the second car has 4x3x2x1 ways it can be loaded.

So the answer should be (I think) 6x5x4x3x4x3x2x1. It's not a combinations question at all, because you said everyone can drive, so each way is a different arrangement of how the people could be in the car.

Or, perhaps you mean it to be a combinations problem, and that's why everyone takes turns driving? In that case, it would be (6x5x4x3x4x3x2x1)/4!(4!) = 15


We have 6 choices for the first car. But won't we have just two choice for the second car. As the other two seats has been booked by Paul and Jess.

Shouldn't the answer be
6x5x4x3x2x1
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 [#permalink] New post 19 Aug 2004, 16:48
Assume Paul drives car 2 and Jess sits shotty.
There are 6! arrangements you can make.


Now, since Paul and jess can occupy seats in car two 4C2 * 2 = 12 ways,


You have 6! * 12 different arrangements.

I would say 8640 ways.
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 [#permalink] New post 19 Aug 2004, 20:12
Got the same result as Ian.
First lets fix car1 (Combination). Six guys have to be arranged in 4 places(Permutation) - Number of ways = 6 x 5x 4 x 3

Now in the remaining two group, P and J join making it four. Four guys have to be arranged in 4 places (Permutation) - Number of ways = 4 x 3 x 2 x 1

Total number of ways = (6 x 5x 4 x 3) x (4 x 3 x 2 x 1) = 8640
  [#permalink] 19 Aug 2004, 20:12
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