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# A committee of 3 people is to be formed from 4 married

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Intern
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A committee of 3 people is to be formed from 4 married [#permalink]

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26 Dec 2006, 19:55
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 formed from 4 married couples. 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 ?

a. 16
b. 24
c. 26
d. 30
e. 32
Intern
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26 Dec 2006, 20:24
Appuvar you got it right ..... OA is E but how did you get to it ?
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26 Dec 2006, 20:25
vmalaiya wrote:
A committee of 3 people is to be formed from 4 married couples. 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 ?

a. 16
b. 24
c. 26
d. 30
e. 32

(AA1) (BB1) (CC1) (DD1)

The first person can be selected 8 different ways.
The second person can be selected 6 different ways. (if you selected A, you can't select A1).
The third person can be selected 4 different ways (if you selected B in second attempt , you can't select B1)

ABC1 and C1AB are same since order doesn't matter.
Therefore number of committees = (8*6*4)/3! =32

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VP
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26 Dec 2006, 20:46
Number of possible combinations of the committee = 8C3 = 56
Number of ways in which one couple can be a part of the committee = 1 * 6

<1 for the couple who will fill 2 seats on the committe. 6 for the ways in which the remaning seat can be filled>

Thus for 4 couples, the number of ways in which a couple can be a part of the committee = 6 * 4 = 24

Thus, number of ways a couple cannot be a part of the committee =
(Total number of combinations) - (no of comb a couple is on the committee)

= 56 - 24 = 32.

Give me E!
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27 Dec 2006, 09:35
I used the same logic that Kripalkavi used.

8C3 - 4*6C1 = 56-24=32!
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27 Dec 2006, 10:12
nice counting everybody....

here is another way to count it, gives the same answer:

first, let choose the three couples (out of 4) that send a representative to the committee... ok there are 4 ways to do that.
now for each couple, independently choose its representative. total of 2 options per couple gives us: 2*2*2=8
multiply the two gives you the right answer.
VP
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27 Dec 2006, 16:24
hobbit wrote:
nice counting everybody....

here is another way to count it, gives the same answer:

first, let choose the three couples (out of 4) that send a representative to the committee... ok there are 4 ways to do that.
now for each couple, independently choose its representative. total of 2 options per couple gives us: 2*2*2=8
multiply the two gives you the right answer.

You COUNT so well hobbit .....Wow thats pretty neat...
Senior Manager
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27 Dec 2006, 17:48
trivikram wrote:
You COUNT so well hobbit .....Wow thats pretty neat...

actually, everybody here did good job as well.... but thanks for the compliment anyway.

there are always so many ways to count things... so there is no right/wrong way, and sometimes what is easy and intuitive for one is difficult for another....

the only principles common to all is: count cautiously. make sure you count every possibility. make sure you don't count a certain possibility more than once. use formulas just if you really sure you need them - i.e. "irreducible" counting tasks... and have fun...

beside that there are just few more "thumb rules" that you learn from experience...

i had once a long post on counting.

anyway.... good counting everybody
VP
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28 Dec 2006, 22:10
do a search. this is posted several times.
28 Dec 2006, 22:10
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