5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
probability
This topic has expert replies
-
Chrystelle
- Junior | Next Rank: 30 Posts
- Posts: 19
- Joined: Thu Dec 07, 2006 3:20 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?
Answer = 20475?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?
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
-
gabriel
- Legendary Member
- Posts: 986
- Joined: Wed Dec 20, 2006 11:07 am
- Location: India
- Thanked: 51 times
- Followed by:1 members
axefx wrote:Answer = 20475?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?
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?gabriel wrote:axefx wrote:Answer = 20475?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?
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
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
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
-
Haldiram Bhujiawala
- Junior | Next Rank: 30 Posts
- Posts: 29
- Joined: Thu Mar 11, 2010 3:23 am
- Thanked: 1 times
-
STEVEN SPIELBERG
- Senior | Next Rank: 100 Posts
- Posts: 43
- Joined: Sat Mar 13, 2010 4:11 am
- Thanked: 1 times
