Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 01 Jul 2016, 18:04

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

David is putting together a golf charity outing. He has

Author Message
Director
Joined: 14 Oct 2003
Posts: 587
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 30 [0], given: 0

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

Show Tags

30 Dec 2003, 22:52
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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: 587
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 30 [0], given: 0

Show Tags

30 Dec 2003, 23:30
Goodness! It's frigg'n 1:30 in the morning - and I'm up writing frigg'n gmat questions - I need counseling!
Senior Manager
Joined: 05 May 2003
Posts: 424
Location: Aus
Followers: 2

Kudos [?]: 10 [0], given: 0

Show Tags

30 Dec 2003, 23:41
Is it ?

9C2 * 8C4 * 4C4 = 2520
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 34

Kudos [?]: 341 [0], given: 0

Show Tags

30 Dec 2003, 23:45
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: 587
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 30 [0], given: 0

Show Tags

30 Dec 2003, 23:54
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: 3461
Followers: 67

Kudos [?]: 810 [0], given: 781

Show Tags

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?

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

Last edited by Praetorian on 31 Dec 2003, 12:46, edited 1 time in total.
CEO
Joined: 15 Aug 2003
Posts: 3461
Followers: 67

Kudos [?]: 810 [0], given: 781

Show Tags

31 Dec 2003, 00:24
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

Director
Joined: 14 Oct 2003
Posts: 587
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 30 [0], given: 0

Show Tags

31 Dec 2003, 07:52
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: 587
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 30 [0], given: 0

Show Tags

31 Dec 2003, 08:05
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?
Senior Manager
Joined: 05 May 2003
Posts: 424
Location: Aus
Followers: 2

Kudos [?]: 10 [0], given: 0

Show Tags

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

The problem says three groups with 4 people.
Director
Joined: 14 Oct 2003
Posts: 587
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 30 [0], given: 0

Show Tags

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

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
Display posts from previous: Sort by