|
Author |
Message |
|
TAGS:
|
|
|
Director
Joined: 14 Oct 2003
Posts: 601
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1
Kudos [?]:
2
[0], given: 0
|
David is putting together a golf charity outing. He has [#permalink]
 30 Dec 2003, 22:52
David is putting together a golf charity outing. He has invited 11 other people and will have to put together 3 foursomes (Group 1, Group 2, Group 3) - one of which he will choose randomly before putting together the rosters. In this group of 11 people are Jacob and Esau. If Jacob and David must be in the same foursome and if Esau cannot be in the same foursome as Jacob (not because he stole Esau's birthright but because Jacob is a slowarse golfer) then how many distinct combination of 3 foursomes can David put together?
|
|
|
|
|
|
|
Director
Joined: 14 Oct 2003
Posts: 601
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Goodness! It's frigg'n 1:30 in the morning - and I'm up writing frigg'n gmat questions - I need counseling! 
|
|
|
|
|
|
Director
Joined: 05 May 2003
Posts: 560
Location: Aus
Followers: 2
Kudos [?]:
3
[0], given: 0
|
Is it ?
9C2 * 8C4 * 4C4 = 2520
|
|
|
|
|
|
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4441
Followers: 10
Kudos [?]:
81
[0], given: 0
|
Did you edit this question? it seemed like I first read "invited 23 other people" instead of 11... I'll give it a harder look tomorrow. gn
_________________
Best Regards,
Paul
|
|
|
|
|
|
Director
Joined: 14 Oct 2003
Posts: 601
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Geethu,
I put down groups 1,2, and 3 as distinct groups. So he chooses to be part of 1 out of the 3 groups first - randomly. Thus 3C1(9C2*8C4*4C4)=7,560. Does this make sense?
Paul, sorry I did edit the question - putting 23 people down would've been too time consuming. Sorry to keep you up!
Titleist
|
|
|
|
|
|
CEO
Joined: 15 Aug 2003
Posts: 3550
Followers: 55
Kudos [?]:
626
[0], given: 781
|
Re: Biblical Golf Combinations [#permalink]
31 Dec 2003, 00:08
Titleist wrote: David is putting together a golf charity outing. He has invited 11 other people and will have to put together 3 foursomes (Group 1, Group 2, Group 3) - one of which he will choose randomly before putting together the rosters. In this group of 11 people are Jacob and Esau. If Jacob and David must be in the same foursome and if Esau cannot be in the same foursome as Jacob (not because he stole Esau's birthright but because Jacob is a slowarse golfer) then how many distinct combination of 3 foursomes can David put together?
EDIT : answer changed
lets say a tentative arrangement looks like this.
GROUP1 - Jacob , David , A ,B
GROUP2- Esau, B ,C,D
GROUP3- No restriction
Two people in group1 can be selected in 9C2 = 36 ways
Three people can be selected in group2 in 7C3 = 35 ways
Four people can be selected in group3 in 4C4 ways =1 ways
Total = 36 * 35 = 1260 ways
each of the these ways result in different groups
answer 1260?
Last edited by Praetorian on 31 Dec 2003, 12:46, edited 1 time in total.
|
|
|
|
|
|
CEO
Joined: 15 Aug 2003
Posts: 3550
Followers: 55
Kudos [?]:
626
[0], given: 781
|
Titleist wrote: Geethu,
I put down groups 1,2, and 3 as distinct groups. So he chooses to be part of 1 out of the 3 groups first - randomly. Thus 3C1(9C2*8C4*4C4)=7,560. Does this make sense?
Paul, sorry I did edit the question - putting 23 people down would've been too time consuming. Sorry to keep you up!
Titleist
could you explain your solution?
|
|
|
|
|
|
Director
Joined: 14 Oct 2003
Posts: 601
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Holy Cow,
Last time I write problems 1:30 in the morning! Kudos Praetorian! I stand corrected. This is the second time in history that Esau has been slighted!
(hold up on the answer! i'm looking it through!)
|
|
|
|
|
|
Director
Joined: 14 Oct 2003
Posts: 601
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Sorry Praetorian!
I came up with a different solution
There's 3C1 ways David can choose a group
For Group 1 (assume David and Jacob are in this Group): (you must take Esau out of the mix; thus it's 8C2) =28
Group 2 (Put Esau back into the mix - no restriction: thus, 8C4)= 70
Group 3 (there are only 3 groups! no restriction: Thus 4C4) = 1
3*28*70*1=5,880
anyone with a different solution?
|
|
|
|
|
|
Director
Joined: 05 May 2003
Posts: 560
Location: Aus
Followers: 2
Kudos [?]:
3
[0], given: 0
|
Re: Biblical Golf Combinations [#permalink]
31 Dec 2003, 08:26
praetorian123 wrote: Titleist wrote: David is putting together a golf charity outing. He has invited 11 other people and will have to put together 3 foursomes (Group 1, Group 2, Group 3) - one of which he will choose randomly before putting together the rosters. In this group of 11 people are Jacob and Esau. If Jacob and David must be in the same foursome and if Esau cannot be in the same foursome as Jacob (not because he stole Esau's birthright but because Jacob is a slowarse golfer) then how many distinct combination of 3 foursomes can David put together? lets say a tentative arrangement looks like this. GROUP1 - Jacob , David , A GROUP2- Esau, B ,C GROUP3- No restriction Group 4 - No restriction the third person in group1 can be selected in 9C1 = 9 ways two people can be selected in group2 in 8C2 = 28 ways three people can be selected in group3 in 6c3 ways =20 ways three people can be selected in group4 in 3c3 = 1 way Total = 9 * 28 * 20 * 1 = 5040 ways each of the these ways result in different groups answer 5040? The problem says three groups with 4 people.
|
|
|
|
|
|
Director
Joined: 14 Oct 2003
Posts: 601
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Re: Biblical Golf Combinations [#permalink]
31 Dec 2003, 08:30
Geethu wrote: praetorian123 wrote: Titleist wrote: David is putting together a golf charity outing. He has invited 11 other people and will have to put together 3 foursomes (Group 1, Group 2, Group 3) - one of which he will choose randomly before putting together the rosters. In this group of 11 people are Jacob and Esau. If Jacob and David must be in the same foursome and if Esau cannot be in the same foursome as Jacob (not because he stole Esau's birthright but because Jacob is a slowarse golfer) then how many distinct combination of 3 foursomes can David put together? lets say a tentative arrangement looks like this. GROUP1 - Jacob , David , A GROUP2- Esau, B ,C GROUP3- No restriction Group 4 - No restriction the third person in group1 can be selected in 9C1 = 9 ways two people can be selected in group2 in 8C2 = 28 ways three people can be selected in group3 in 6c3 ways =20 ways three people can be selected in group4 in 3c3 = 1 way Total = 9 * 28 * 20 * 1 = 5040 ways each of the these ways result in different groups answer 5040? The problem says three groups with 4 people. Yes the question does state three groups with 4 people.
|
|
|
|
|
|
|
Re: Biblical Golf Combinations
[#permalink]
31 Dec 2003, 08:30
|
|
|
|
|
|
|
|
|
|
|
close
