2 Combination Qs. 1. How many ways to choose a group of 3 : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 01:53

### 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

# 2 Combination Qs. 1. How many ways to choose a group of 3

Author Message
Intern
Joined: 29 Jul 2005
Posts: 36
Followers: 0

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

2 Combination Qs. 1. How many ways to choose a group of 3 [#permalink]

### Show Tags

18 Aug 2005, 01:43
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

2 Combination Qs.

1. How many ways to choose a group of 3 people from 4 couples so that no couples are chosen?

2. A box contains 5 diff. red and 6 diff. white balls. In how many ways can 6 balls be selected so that there are atleast 2 balls of each color?

Thanks.
Senior Manager
Joined: 30 May 2005
Posts: 373
Followers: 1

Kudos [?]: 11 [0], given: 0

### Show Tags

18 Aug 2005, 02:53
Priti wrote:
2 Combination Qs.

1. How many ways to choose a group of 3 people from 4 couples so that no couples are chosen?

Select 3 couples and then choose 1 from each.

C(4,3) * C(2,1) * C(2,1) * C(2,1) = 4 * 2 * 2 * 2= 32
Senior Manager
Joined: 30 May 2005
Posts: 373
Followers: 1

Kudos [?]: 11 [0], given: 0

### Show Tags

18 Aug 2005, 02:58
Quote:
2. A box contains 5 diff. red and 6 diff. white balls. In how many ways can 6 balls be selected so that there are atleast 2 balls of each color?

We need to sum up mutually exclusive combos.

2 of one kind, 4 of another ---- A
3 each --- B

A = C(5,2) * C (6,4) + C (5,4) + C (6,2) = 10 * 15 + 5 * 15 = 225
B = C (5,3) * C (6,3) = 10 * 20 = 200

Total = 425
Manager
Joined: 14 Jul 2005
Posts: 104
Location: Sofia, Bulgaria
Followers: 1

Kudos [?]: 9 [0], given: 0

### Show Tags

18 Aug 2005, 04:57
1) I couldn't come up with a quick solution for the first problem without looking at ajb's post. I agree with him on the approach.

2) I did basically the same thing as ajb, only the other way around. I considered

(All possibilities) - (0 red, six white) - (1 red, 5 white) - (1 white, 5 red) =
C(6,11) - C(0,5)*C(6,6) - C(1,5)*C(5,6) - C(1,6)*C(5,5) = 425
Intern
Joined: 13 Aug 2005
Posts: 26
Location: Israel
Followers: 0

Kudos [?]: 14 [0], given: 0

### Show Tags

18 Aug 2005, 05:17
Priti wrote:
2 Combination Qs.

1. How many ways to choose a group of 3 people from 4 couples so that no couples are chosen?

Total combinations of choosing 3 out of 8 (order not important): 8!/(5!*3!) = 56
Combinations for groups with couples: 4 couples * 6 (one from each other couple) = 24
Total combinations - Combinations for groups with couples = 32
Director
Joined: 03 Nov 2004
Posts: 865
Followers: 2

Kudos [?]: 54 [0], given: 0

### Show Tags

18 Aug 2005, 05:27
1. Three people can be chosen from a group of 8(4 couples) in 8C3 ways = 56 ways
Number of ways to choose 1 couple from a group of 4 is 4 * 6 - after choosing 1 couple there are 6 different ways to choose the other person.
So number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 = 32 ways.

2.
2R + 4W = 5C2 * 6C4 = 10 * 15 = 150
4R + 2W = 5C4 * 6C2 = 5 * 15 = 75
3R + 3W = 5C3 * 6C3 = 10 * 20 = 200
which is 425 ways
Senior Manager
Joined: 30 May 2005
Posts: 276
Followers: 1

Kudos [?]: 35 [0], given: 0

### Show Tags

18 Aug 2005, 08:52
1. Three people can be chosen from a group of 8(4 couples) in 8C3 ways = 56 ways
Number of ways to choose 1 couple from a group of 4 is 4 * 6 - after choosing 1 couple there are 6 different ways to choose the other person.
So number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 = 32 ways.

2.
2R + 4W = 5C2 * 6C4 = 10 * 15 = 150
4R + 2W = 5C4 * 6C2 = 5 * 15 = 75
3R + 3W = 5C3 * 6C3 = 10 * 20 = 200
which is 425 ways

plz rthothad i don 't understand why you do retrieve 24 from 56; Plz can you explain why number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 thanks

regards

mandy
Senior Manager
Joined: 29 Nov 2004
Posts: 484
Location: Chicago
Followers: 1

Kudos [?]: 25 [0], given: 0

### Show Tags

18 Aug 2005, 09:28
Q1)

8*6*4/3! = 32 (8 choices for first, 6 for second, 4 for 3rd and divide by 6 to get rid of duplicates, since order does not matter)

Q2) 425, Good explanations above
_________________

Fear Mediocrity, Respect Ignorance

Director
Joined: 03 Nov 2004
Posts: 865
Followers: 2

Kudos [?]: 54 [0], given: 0

### Show Tags

18 Aug 2005, 18:44
mandy wrote:
plz rthothad i don 't understand why you do retrieve 24 from 56; Plz can you explain why number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 thanks

regards

mandy

mandy, My initial 8C3 combination contains couples also, so in order to remove all the combination that contains a couple, I had to deduct. Hope I am clear.
18 Aug 2005, 18:44
Display posts from previous: Sort by