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

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A committee of 3 people is to be chosen from four married [#permalink]

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01 Aug 2006, 05:11
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A committee of 3 people is to be chosen from four married couple. What is the number of different committees that can be chosen if 2 people who are married to each other cannot both serve on the committee.

1. 16
2. 24
3. 26
4. 30.
5. 32
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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01 Aug 2006, 09:25
E
Number of ways of choosing first person = 8
Number of ways of choosing second person = 6 (Out of 7 remaning, anyone can be chosen except his/her partner)
Number of ways of choosing third person = 4 (Out of 6 remaning anyone can be chosen except partners of two selected person)

Total = 8*6*4
But this contain duplicates. To remove duplicates divide by the factorial of number of people in the committee.
So total ways = 8*6*4/3! = 32
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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01 Aug 2006, 09:42
Thanks for the answer. Can this question also be solved using the xCn (combination) formula?
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01 Aug 2006, 10:02
Yes, that is:

8C3-4*6C1

Where, 8C3 is the different ways you can choose 3 persons from the 8 possibilities.
PPX=groups where the partners are together..so that is 4*6=24

56-24=32
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02 Aug 2006, 03:51
E 32

Number of ways 3 people can be chosen from 8 people = 8C3 = 56

When the fist married couple is chosen .... number of ways to choose the third person = 6
Similarily since there are four couples.... number of ways to choose couple in the commitee = 4*6 = 24

Hence number of ways NOT to choose any couple = 56-24 = 32
02 Aug 2006, 03:51
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