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David is putting together a golf charity outing. He has [#permalink]

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30 Dec 2003, 21:52

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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?

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
_________________

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!

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, 11:46, edited 1 time in total.

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!

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

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.

gmatclubot

Re: Biblical Golf Combinations
[#permalink]
31 Dec 2003, 07:30