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A team of 8 students goes on an excursion, in two cars, of

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CEO
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A team of 8 students goes on an excursion, in two cars, of [#permalink] New post 07 Oct 2003, 19:44
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A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

1) 9
2) 26
3) 126
4) 3920


please explain
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 [#permalink] New post 07 Oct 2003, 22:38
8=4+4 => 8C4=70, I would divide 70 by 2 because doubling effecr is here
8=5+3 => 8C5=56

70+56=126
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Re: PS : Counting Methods ( Two Cars , 8 Students) [#permalink] New post 08 Oct 2003, 16:30
praetorian123 wrote:
A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

1) 9
2) 26
3) 126
4) 3920


please explain


we have two cases

1. divide 8 into 5 and 3

# of ways = 8! /(5!*3!) = 56

2. divide 8 into 4 and 4

# of ways = 8! / (4! *4!) = 70

total = > 126

stolyar, i dont think we need to divide the 70 by 2...
for example, if you have ABCD and EFGH ...both of these groups when selected will have unique combinations...

thanks
praetorian
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 [#permalink] New post 15 Oct 2003, 17:25
if we have one car with 5 spaces and the other with 3 spaces for 8 people, I would you say that there are 70 diferent ways to travel, not 140, right? Because the 3 left from the 5-space car were already counted in the 3-space car.
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 [#permalink] New post 16 Oct 2003, 01:15
Yoda wrote:
if we have one car with 5 spaces and the other with 3 spaces for 8 people, I would you say that there are 70 diferent ways to travel, not 140, right? Because the 3 left from the 5-space car were already counted in the 3-space car.



dont understand what you are trying to say.

do you agree with my answer?
  [#permalink] 16 Oct 2003, 01:15
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