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# A group of 10 people consists of 3 married couples and 4

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Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 641
A group of 10 people consists of 3 married couples and 4  [#permalink]

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17 May 2011, 08:18
3
1
19
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(N/A)

Question Stats:

0% (00:00) correct 100% (00:49) wrong based on 30 sessions

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A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

OA is 207.

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method
The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method
---------------
No of ways to select one couple and two singles + No of ways to select all singles
Senior Manager
Joined: 24 Mar 2011
Posts: 290
Location: Texas
Re: A group of 10 people  [#permalink]

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17 May 2011, 09:27
Complicated method -

You have to consider all the possibilities.

Obvious cases -
All 4 singles &
1 married couple and 2 singles

Other cases -
1 marrried person from a couple and 3 singles
1 married couple, 1 married person from a couple and 1 single
1 married couple, 2 married person from different couples
1 married person from each couple and 1 single
1 married person from 2 separate couples and 2 singles

i still dont know if i missing any other case...
Director
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Re: A group of 10 people  [#permalink]

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17 May 2011, 11:23
well the approach has been discussed over here.

permutation-and-combination-help-110982.html

hope this will help.
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
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Re: A group of 10 people  [#permalink]

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17 May 2011, 18:59
fluke, Ian, Karishma anyone?

This is how I tried counting it forward.

1 couple and 3 singles
---------------------
One couple, 2 singles among rest of couples, 1 singles from 4
3C1 * (6 * 4/2) * 4C1 = 3 * 12 * 4 = 144

One couple, 1 single among rest of couples, 2 singles from 4
3C1 * 6 * 4C2 = 3 * 6 * 6 = 108

One couple, 3 singles from 4
3C1 * 4C3 = 3 * 4 = 12

All 4 singles
-------------
3 singles among 3 couples, 1 single from 4
(6 * 4 * 2) / 3! * 4C1 = 8 * 4 = 32

2 singles among 3 couples, 2 single from 4
(6 * 4)/2 * 4C2 = 12 * 6 = 72

1 single among 3 couples, 3 single from 4
6 * 4C3 = 24

4 singles from 4
4C4 = 1

Adding all the above = 264+133 = 397. Way tooo much than the answer. Can you point the error?
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 641
Re: A group of 10 people  [#permalink]

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17 May 2011, 21:06
I may be drunk that's why. Thanks so much ! yes it all adds up to sanity i.e. 207

VeritasPrepKarishma wrote:
gmat1220 wrote:
fluke, Ian, Karishma anyone?

This is how I tried counting it forward.

1 couple and 3 singles - C'mon! 1 couple (2 people) + 3 singles? That's 5 people!
---------------------
One couple, 2 singles among rest of couples, 1 singles from 4
3C1 * (6 * 4/2) * 4C1 = 3 * 12 * 4 = 144

One couple, 1 single among rest of couples, 2 singles from 4
3C1 * 6 * 4C2 = 3 * 6 * 6 = 108

One couple, 3 singles from 4
3C1 * 4C3 = 3 * 4 = 12

All 4 singles - Looks Fine
-------------
3 singles among 3 couples, 1 single from 4
(6 * 4 * 2) / 3! * 4C1 = 8 * 4 = 32

2 singles among 3 couples, 2 single from 4
(6 * 4)/2 * 4C2 = 12 * 6 = 72

1 single among 3 couples, 3 single from 4
6 * 4C3 = 24

4 singles from 4
4C4 = 1

Adding all the above = 264+133 = 397. Way tooo much than the answer. Can you point the error?
Current Student
Joined: 26 May 2005
Posts: 443
Re: A group of 10 people  [#permalink]

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17 May 2011, 21:50
2
gmat1220 wrote:
A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

OA is 207.

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method
The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method
---------------
No of ways to select one couple and two singles + No of ways to select all singles

Let A----B
C----D
E ----F are couples and
1
2
3
4 are singles

1st scenario .. 4 singles out of 4 singles ------- 4c4 = 1
2nd scenario 2 Married singles + 2 singles ------ ( 6*4)/2 * 4c2 = 72
3rd 1 Married single + 3 singles ------- 6*4c3 = 24
4th 3 married single + 1 single ------- (6*4*2)/3!*4C1 = 32
5th scenario 1Couple + 1 Married single + 1 single ---- 3c1*4*4c1= 48
6th scenario 1 couple + 2 Married singles ------- 3c1*( 4*2)/2 = 12
7th scenario 1 couple + 2 singles --------- 3c1*4C2 = 18

Total = 1+72+24+ 32+48+12+18 = 207
Intern
Joined: 12 May 2011
Posts: 2
Re: A group of 10 people  [#permalink]

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14 Sep 2011, 13:24
1
1
sudhir18n wrote:
gmat1220 wrote:
A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

OA is 207.

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method
The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method
---------------
No of ways to select one couple and two singles + No of ways to select all singles

Let A----B
C----D
E ----F are couples and
1
2
3
4 are singles

1st scenario .. 4 singles out of 4 singles ------- 4c4 = 1
2nd scenario 2 Married singles + 2 singles ------ ( 6*4)/2 * 4c2 = 72
3rd 1 Married single + 3 singles ------- 6*4c3 = 24
4th 3 married single + 1 single ------- (6*4*2)/3!*4C1 = 32
5th scenario 1Couple + 1 Married single + 1 single ---- 3c1*4*4c1= 48
6th scenario 1 couple + 2 Married singles ------- 3c1*( 4*2)/2 = 12
7th scenario 1 couple + 2 singles --------- 3c1*4C2 = 18

Total = 1+72+24+ 32+48+12+18 = 207

hi - you could have simply done this ..

Total number of ways of selecting 4 people out of 10 = 10C4
Total number of ways of selecting 2 married couple = 3C2

Answer = 10C4 - 3C2 = 210-3 = 207..
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Re: A group of 10 people  [#permalink]

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10 Oct 2011, 06:32
1
total number of combinations 10c4 = 10!/4!6! = 210
total number of married couple = 3c2 = 3!/2! = 3
total number of possible commities are 207
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Re: A group of 10 people  [#permalink]

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10 Oct 2011, 07:39
if we change the question to "...at least 1 married couple", whats gonna be the answer?
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Joined: 18 Mar 2011
Posts: 40
Schools: Foster '17 (A)
Re: A group of 10 people  [#permalink]

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11 Oct 2011, 23:36
1
If its atleast one couple.. then shud be 3C1*4C2 + 3C2 ways = 21 ways
Manager
Joined: 19 Oct 2011
Posts: 58
Re: A group of 10 people  [#permalink]

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16 Nov 2011, 16:38
We can use the other way of thinking, that is (total possibility of selecting 4 people)-(selecting two couples)
then the answer will be 10C4-3C2=210-3=207. That's much more quickly
Manager
Joined: 07 Feb 2011
Posts: 89
Re: A group of 10 people  [#permalink]

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13 Feb 2013, 18:27
One way to do problem: Answer = No of ways to select one couple and two singles + No of ways to select all singles=207=official answer

"The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207"

Since when has the opposite of "at most 1" equaled to exact no of ways to select 2 couples? Could someone help me grasp why this is conceptually the opposite of exactly 1 couple?
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Joined: 16 Oct 2010
Posts: 9874
Location: Pune, India
Re: A group of 10 people  [#permalink]

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13 Feb 2013, 20:31
1
2
manimgoindowndown wrote:
One way to do problem: Answer = No of ways to select one couple and two singles + No of ways to select all singles=207=official answer

"The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207"

Since when has the opposite of "at most 1" equaled to exact no of ways to select 2 couples? Could someone help me grasp why this is conceptually the opposite of exactly 1 couple?

You need to select a committee of 4. You can do it in 3 ways:
a. Select no couple and 4 others
b. Select one couple and 2 others
c. Select two couples

'At most one' means either no couples or only one couple. What is the complement of at most one? The only other way in which you can select a committee of 4 is by selecting 2 couples. So you can find the number of ways of selecting 2 couples and subtract that from the total number of ways of selecting 4 people. You will get the number of ways in which there will be at most one couple.
_________________
Karishma
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Joined: 07 Feb 2011
Posts: 89
Re: A group of 10 people  [#permalink]

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15 Feb 2013, 04:47
VeritasPrepKarishma wrote:
manimgoindowndown wrote:
One way to do problem: Answer = No of ways to select one couple and two singles + No of ways to select all singles=207=official answer

"The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207"

Since when has the opposite of "at most 1" equaled to exact no of ways to select 2 couples? Could someone help me grasp why this is conceptually the opposite of exactly 1 couple?

You need to select a committee of 4. You can do it in 3 ways:
a. Select no couple and 4 others
b. Select one couple and 2 others
c. Select two couples

'At most one' means either no couples or only one couple. What is the complement of at most one? The only other way in which you can select a committee of 4 is by selecting 2 couples. So you can find the number of ways of selecting 2 couples and subtract that from the total number of ways of selecting 4 people. You will get the number of ways in which there will be at most one couple.

Namaste aur shukria so much again Karishma. Made it clear within the context of the problem
Math Expert
Joined: 02 Sep 2009
Posts: 59712
Re: A group of 10 people consists of 3 married couples and 4  [#permalink]

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16 Feb 2013, 04:10
gmat1220 wrote:
A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

OA is 207.

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method
The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method
---------------
No of ways to select one couple and two singles + No of ways to select all singles

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Hope it helps.
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Re: A group of 10 people consists of 3 married couples and 4  [#permalink]

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22 Feb 2013, 07:47
my question is :
i understand we can do like selecting 2 couples and subtract that from the total number of ways of selecting 4 people.
now there 6 married peple and 4 singles
so it should be 10c4 - 6x5x4x3/4x3x2x1

but i dont understand how 10c4- 3x2/2 comes?? can someone please explain.
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Re: A group of 10 people consists of 3 married couples and 4  [#permalink]

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22 Feb 2013, 08:04
1
2
skamal7 wrote:
my question is :
i understand we can do like selecting 2 couples and subtract that from the total number of ways of selecting 4 people.
now there 6 married peple and 4 singles
so it should be 10c4 - 6x5x4x3/4x3x2x1

but i dont understand how 10c4- 3x2/2 comes?? can someone please explain.

We have 3 couples, so 3C2=3!/2!=3 is the number of ways to choose 2 couples out of 3.

Hope it's clear.
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Re: A group of 10 people consists of 3 married couples and 4  [#permalink]

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22 Feb 2013, 08:10
bunnel thanks for kind response.

One thing i dont understand is while taking total number of ways of selecting 4 people we count the 3 married couples as 6 right.?so when we are selecting 2 married couples it should be 6x5x4x3/4! na??but why when selecting 2 married couples we are taking count just as 3x2/2!?? please do explain!!!
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Posts: 59712
Re: A group of 10 people consists of 3 married couples and 4  [#permalink]

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22 Feb 2013, 08:13
1
skamal7 wrote:
bunnel thanks for kind response.

One thing i dont understand is while taking total number of ways of selecting 4 people we count the 3 married couples as 6 right.?so when we are selecting 2 married couples it should be 6x5x4x3/4! na??but why when selecting 2 married couples we are taking count just as 3x2/2!?? please do explain!!!

We are interested in number of ways we can select 2 couples out of 3, so 2 entities out of 3.
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Re: A group of 10 people consists of 3 married couples and 4  [#permalink]

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Re: A group of 10 people consists of 3 married couples and 4   [#permalink] 29 Jun 2019, 14:51
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