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probability

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by Chrystelle » Fri Jan 12, 2007 7:03 am
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
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Re: probability

by axefx » Wed Jan 17, 2007 1:50 pm
Chrystelle wrote:A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
Answer = 20475?

possible teams:

12,13,14,15,16,17,18
23,24,25,26,27,28
34,35,36,37,38
45,46,47,48
56,57,58
67,68
78

7 + 6 + 5 + 4 + 3 + 2 + 1 = 28

choosing 4 teams out of 28 will be (28 C 4) =

28! / 4! * (28 - 4)!
= 28 * 27 * 26 * 25 * 24! / 4! * 24!
= 28 * 27 * 26 * 25 / 4 * 3 * 2 * 1
= 20475
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Re: probability

by gabriel » Fri Jan 19, 2007 11:39 am
axefx wrote:
Chrystelle wrote:A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
Answer = 20475?

possible teams:

12,13,14,15,16,17,18
23,24,25,26,27,28
34,35,36,37,38
45,46,47,48
56,57,58
67,68
78

7 + 6 + 5 + 4 + 3 + 2 + 1 = 28

choosing 4 teams out of 28 will be (28 C 4) =

28! / 4! * (28 - 4)!
= 28 * 27 * 26 * 25 * 24! / 4! * 24!
= 28 * 27 * 26 * 25 / 4 * 3 * 2 * 1
= 20475

hmm... that is wrong by a long shot..... what if u select team `12 & 13 how are they going to play double tennis with 1 being in both teams....

the correct answer is 8!/( 2!*2!*2!*2!*4!)= 105
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Re: probability

by axefx » Fri Jan 26, 2007 10:38 am
gabriel wrote:
axefx wrote:
Chrystelle wrote:A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
Answer = 20475?

possible teams:

12,13,14,15,16,17,18
23,24,25,26,27,28
34,35,36,37,38
45,46,47,48
56,57,58
67,68
78

7 + 6 + 5 + 4 + 3 + 2 + 1 = 28

choosing 4 teams out of 28 will be (28 C 4) =

28! / 4! * (28 - 4)!
= 28 * 27 * 26 * 25 * 24! / 4! * 24!
= 28 * 27 * 26 * 25 / 4 * 3 * 2 * 1
= 20475

hmm... that is wrong by a long shot..... what if u select team `12 & 13 how are they going to play double tennis with 1 being in both teams....

the correct answer is 8!/( 2!*2!*2!*2!*4!)= 105
Good point, which I entirely missed. Can you please provide explanations for your analysis?
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2520???

by yuseop » Sat Jan 27, 2007 11:00 pm
Isn't it simply 8C2 * 6C2 * 4*2?
Yuseop
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2520???

by yuseop » Sat Jan 27, 2007 11:33 pm
Isn't it simply 8C2 * 6C2 * 4*2?
Yuseop
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by rs123 » Tue Jan 30, 2007 9:00 am
It is simply

8C2 * 6C2 * 4C2 * 2C2 / 4!

8!/(2! * 2! * 2! * 2! * 4!)

8C2 * 6C2 * 4C2 * 2C2 means you keep on selecting 2 at a time. We need to divide by 4! because a combination like AB, CD, EF, GH could be permuated among itself in 4! ways.

HTH
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by Haldiram Bhujiawala » Fri May 21, 2010 8:27 am
IMO[/spoiler]105[spoiler][/spoiler]
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by STEVEN SPIELBERG » Sat May 22, 2010 7:15 am
IMO105
I want to win an OSCAR on the GMAT !!!
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