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gmatprep- married couples

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by bkobilov » Sun Oct 03, 2010 10:35 am
...

This question might be asked before, but i cannot remember the solution.

A committee of three people is to be chosen from four married couples. What is the number of different commities that can be chosen if two people who are married to each other cannot both serve on this committee?

16
24
26
30
32

OA: E
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by shovan85 » Sun Oct 03, 2010 12:09 pm
3 places in committee out of 8 people (4 couples) can be chosen in 8C3 ways = 56 ways.

Now say every time you choose a group of committee a couple is bound to be selected. Thus the number of ways this is possible
= (Number of ways 2 places can be selected from 4 couples) * (Number of ways remaining 1 place of committee is selected from 86 people [as one couple is already selected])
= (4C1) * (6C1)
= 24

Thus, Required answer is 56-24 = 32
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by shovan85 » Sun Oct 03, 2010 12:10 pm
shovan85 wrote:3 places in committee out of 8 people (4 couples) can be chosen in 8C3 ways = 56 ways.

Now say every time you choose a group of committee a couple is bound to be selected. Thus the number of ways this is possible
= (Number of ways 2 places can be selected from 4 couples) * (Number of ways remaining 1 place of committee is selected from 86 people [as one couple is already selected])
= (4C1) * (6C1)
= 24

Thus, Required answer is 56-24 = 32
Typo its 6 not 86 :)
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by GMATGuruNY » Sun Oct 03, 2010 5:14 pm
bkobilov wrote:...

This question might be asked before, but i cannot remember the solution.

A committee of three people is to be chosen from four married couples. What is the number of different commities that can be chosen if two people who are married to each other cannot both serve on this committee?

16
24
26
30
32

OA: E
Another approach:

Number of options for the 1st person = 8.
Number of options for the 2nd person = 6. (Of the 7 people left, we can't use the mate of the 1st person chosen, leaving 7-1= 6 choices.)
Number of options for the 3rd person = 4. (Of the 6 people left, we can't use the mates of the 2 people already chosen, leaving 6-2 = 4 choices.)

Since when we choose a COMMITTEE the ORDER of the selections doesn't matter, we divide by the number of ways to arrange the 3 people chosen:
(8*6*4)/3! = 32.

The correct answer is E.
Last edited by GMATGuruNY on Sun Sep 09, 2012 2:45 am, edited 1 time in total.
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by pzazz12 » Mon Oct 04, 2010 4:35 am
shovan85 wrote:3 places in committee out of 8 people (4 couples) can be chosen in 8C3 ways = 56 ways.

Now say every time you choose a group of committee a couple is bound to be selected. Thus the number of ways this is possible
= (Number of ways 2 places can be selected from 4 couples) * (Number of ways remaining 1 place of committee is selected from 86 people [as one couple is already selected])
= (4C1) * (6C1)
= 24

Thus, Required answer is 56-24 = 32


thank you..........know i understood how to solve this type of problems is there any alternative method to solve.......
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by shovan85 » Mon Oct 04, 2010 6:20 am
pzazz12 wrote:
shovan85 wrote:3 places in committee out of 8 people (4 couples) can be chosen in 8C3 ways = 56 ways.

Now say every time you choose a group of committee a couple is bound to be selected. Thus the number of ways this is possible
= (Number of ways 2 places can be selected from 4 couples) * (Number of ways remaining 1 place of committee is selected from 86 people [as one couple is already selected])
= (4C1) * (6C1)
= 24

Thus, Required answer is 56-24 = 32


thank you..........know i understood how to solve this type of problems is there any alternative method to solve.......
Alternate method is shown by GMATGuruNY. But honestly if you break it stepwise and solve then you will have a better hold and visualization on the problem. See what fits you best.
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