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A committee of 3 people is to be chosen from four married

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Re: A committee of 3 people is to be chosen from four married [#permalink] New post 07 Nov 2012, 04:37
Expert's post
watwazdaquestion wrote:
is this a correct way to get the answer? or was it just coincidence:

8C3 - 4(4C2) = 56 - 4(6) = 32


It's not clear what is the logic behind the formula.

Reversed approach would be:
There are 8C3=56 ways to select 3 people out of 8 without any restriction;
There are 4C1*6=24 ways there to be a couple among 3 members: 4C1 ways to select a couple out of 4, which will be in the committee and 6 ways to select the third remaining member (since there will be 6 members left after we select a couple out of 8 people).

56-24=32.

Hope it's clear.
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Re: A committee of 3 people is to be chosen from four married [#permalink] New post 27 Dec 2012, 20:54
LM wrote:
A committee of 3 people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A. 16
B. 24
C. 26
D. 30
E. 32



How many ways to select 3 represented couples from 4 couples? 4!/3!1! = 4
How many ways to select a person from a pair? 2
=4 * 2 * 2 * 2 = 32

Answer: E

More detailed explanation here : Selection/Deselection Technique
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Re: A committee of 3 people is to be chosen from four married [#permalink] New post 02 Sep 2013, 22:55
I solved it using this method, hope I'm using the correct concept

total 4 couples = 8 people in total
total no of ways to choose 3 people out of 8 = 8!/(5!3!) = 56
No. of ways couples are included in the com = 4! = 24
Therefore no. of couples with no couples included = 56-24 = 32.
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Problem Solving Question [#permalink] New post 26 Oct 2013, 15:33
A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A. 16
B. 24
C. 26
D. 30
E. 32

Please explain the answer.
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Re: Problem Solving Question [#permalink] New post 26 Oct 2013, 18:03
schokshi99 wrote:
A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A. 16
B. 24
C. 26
D. 30
E. 32

Please explain the answer.

Because people who are married to each other cannot both serve on the committee, three members of the committee must come from 3 different couples. We have 4 ways to pick 3 couples from 4 couples. For each couple, we have 2 choices to pick one of them.
So the answer is 4 * 2 * 2 *2 = 32 (E)

Last edited by tuanle on 26 Oct 2013, 20:39, edited 2 times in total.
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Re: Problem Solving Question [#permalink] New post 26 Oct 2013, 20:20
Agreed with Tuanle.

Another easy way is to write down all the ways to select

We have total 4 men and 4 women from which we need to form a committee of 3 such that husband and wife wont be included

The combinations are

1) MMM (all 3 men) - 4c3 = 4
2) MMW (Two men and one women) - 4c2 * 2c1 (the wives of selected men should not be included) = 12
3) WWM - 4c2*2c1 = 12
4) WWW - 4c3 = 4

Total = 4+12+12+4=32
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Re: Problem Solving Question [#permalink] New post 27 Oct 2013, 04:47
Expert's post
schokshi99 wrote:
A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A. 16
B. 24
C. 26
D. 30
E. 32

Please explain the answer.


Merging similar topics. Please refer to the solutions provided.

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Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rule 1 and 3. Thank you.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: A committee of 3 people is to be chosen from four married [#permalink] New post 12 Nov 2013, 10:42
instead of thinking as husband and wife in the group, think of them as all different ppl in groups of two. you can only select one person from each group

so that means

(a,b) (c,d) (e,f) (g,h)

and we need to fill in 3 spaces

_ x_ x_

any one person can come from each group, so three spaces filled by one person from each group are

2x2x2 - (A)

taking different combos of the group => 4C3 => 4 - (B)

multiply (A) and (B) gives all the different ways this group can be created

(A) x (B) = 2x2x2x4 = 32 Thus choice E
Re: A committee of 3 people is to be chosen from four married   [#permalink] 12 Nov 2013, 10:42
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