GMAT Prep Test

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GMAT Prep Test

by awilhelm » Thu Dec 18, 2008 6:05 pm
Please help...

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 can't both serve on the committee?

a) 16
b) 24
c) 26
d) 30
e) 32
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by logitech » Thu Dec 18, 2008 6:30 pm
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by VP_Jim » Fri Dec 19, 2008 7:48 am
I like to get away from using formulas on permutation/combination problems. Most GMAT problems are easier if you just think about them logically. In this problem, we have three "slots" on the team, so draw a little picture:

_____ ______ ______

And ask yourself: how many choices do I have for each slot?

For the first slot, you can pick anyone, so 8. For the second slot, you can pick anyone except the person you already picked and his or her spouse, so 6. For the last slot, using the same reasoning, we have 4 choices.

So: 8x6x4

But, since order doesn't matter in this problem, we have to divide by 3x2x1 (an easy way to think of it is to divide by N!, where "N" is the number of items in the numerator of your fraction).

(8x6x4)/(3x2x1) = 32
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by lunarpower » Sat Dec 20, 2008 2:35 am
you can also calculate this indirectly.

there are only 2 possibilities:
(a) nobody on the committee is married to anybody else on the committee;
(b) exactly one married couple is on the committee.

we want (a). therefore, we can also find it by finding the total number of possible committees, and subtracting out (b).

the total number of committees is 8c3 = 8!/(3!5!), or 56.
you can also calculate this number as (8x7x6)/(6!). remember that the multiplication rule always presupposes that "order matters" (i.e., the different positions are distinguishable), so here we have to divide by 3! to negate that assumption.

for (b), calculating the total # of committees involving one of the married couples: just take the number of couples (4) times the number of possible 3rd persons (6). that's 4 x 6 = 24, so 56 - 24 = 32.

56 - 32 = 24.

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in case any poster is going to ask the inevitable "which of these solutions is better?"-type question, the answer to that question is, as usual, "whichever one you think of first".
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Not Sure

by CrackGMAC » Thu Dec 25, 2008 6:31 am
Hi Ron,
I followed exactly the same method as yours but I have two issues in calc part b.

First, I made it 5*6 = 30 instead of 4 x 6 because position 1 can be filled by any one person of the 5 married couples and next position has to be filled in one ways then the last position can be filled in 6 ways which makes it 30 instead of 24.

Secondly, I am not sure about this because you can choose A1 A2 * 6 or A2 A1 * 6 which makes 5 * 2 * 6. Is it because arrangement does not matter its only the selection and hence it 56 - 5 * 6 = 26

Whats the OA awilhelm
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Re: Not Sure

by lunarpower » Wed Dec 31, 2008 5:37 am
one:
First, I made it 5*6 = 30 instead of 4 x 6 because position 1 can be filled by any one person of the 5 married couples and next position has to be filled in one ways then the last position can be filled in 6 ways which makes it 30 instead of 24.
huh?

per the problem statement, there are four, not five, married couples.
go back and read the problem statement again.

two:
Secondly, I am not sure about this because you can choose A1 A2 * 6 or A2 A1 * 6 which makes 5 * 2 * 6.
no, you don't multiply this by two.

analogy #1:
if you have 3 shirts and 4 pairs of pants, then you have 3 x 4 = 12 outfits.
hopefully you consider that "obvious"; the multiplication principle at work here is probably the most basic principle in all of combinatorics.
the point, though, is that your approach above would give 3 x 4 x 2 = 24 outfits, an incorrect answer.

analogy #2:
there are 26 letters in the alphabet.
this question is sort of like saying: 'if i count all the letters in the alphabet forwards, and then i count them all backwards, i get a total of 52. so why aren't there 52 letters?'
answer:
once you've figured out one legitimate way to count something, you just count it that way. if you count the same things twice in two different ways, then you'll calculate twice the number you're supposed to get.

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now, if the ORDER in which you pick the couple and the lone individual mattered - i.e., if couple/individual differs in some way from individual/couple - then your approach would be correct.
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Thanks Ron

by CrackGMAC » Thu Jan 01, 2009 5:15 am
:D

Thanks Ron that helped. Can you suggest anyways in which I can improve in following topics in qant. If you have some questions

Set Theory
Stats
Number System
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by maihuna » Thu Jan 01, 2009 6:37 am
I always see this question in following ways:

M1 M3 M3 M4
F1 F2 F3 F4

Assuming M1 F1 are the married pair.

I can do the cal in following way:

Total way of choosing 3 out of 8 is 8c3 = 8x7x6/3x2 = 56

How many couples can be there: At most 1, that is the key here, because only 3 people are there: In how many ways a non-desired combo can be there:

Choose a cople: In one of 4 ways, for all such any of the remaining 6 can be choosen so a total of 4x6 = 24 ways

Desired = Total - Not Desired = 56 - 24 = 32

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Re: Thanks Ron

by lunarpower » Thu Jan 01, 2009 2:38 pm
CrackGMAC wrote::D

Thanks Ron that helped. Can you suggest anyways in which I can improve in following topics in qant. If you have some questions

Set Theory
Stats
Number System
hmm.

well, those are some awfully general topics. do you have any more specific sub-areas on which you want to improve?

when you say "set theory", are you talking about overlapping sets?

as far as statistics, do you have problems with the mean? median? all of the above?

i'm not sure what you mean by "number system".

in any case, the best bet is usually to post specific problems, rather than general subject areas such as the ones you've mentioned. even if you're posting particular problems, we can still include general takeaways in our answers to those posts.
Ron has been teaching various standardized tests for 20 years.

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Questions on Stats

by CrackGMAC » Sat Jan 03, 2009 12:05 am
Ya in set theory, I more concerned abt overlapping sets problem and Std Deviation and Normal Deviation in Stats.

I am browsing the site but its diff to find good set of question. If you have few and post for ref. It will be of grt help.

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