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by prachi18oct » Fri May 08, 2015 12:33 pm
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What is wrong with the below solution?

Number of ways we can select first member = 10C1
Number of ways we can select second member = 8C1 ( 1 already selected and so his/her partner excluded)
Number of ways we can select third member = 6C1 ( 2 already selected and so their partner excluded)
10 * 8 * 6 = 480

Please explain!
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by Brent@GMATPrepNow » Fri May 08, 2015 12:41 pm
If a committee of 3 people is to be selected from among 5 married couples so that the
committee does not include two people who are married to each other, how many such
committees are possible?
A. 20
B. 40
C. 50
D. 80
E. 120

Take the task of creating a committee and break it into stages.

Stage 1: Select 3 COUPLES
Since the order in which we select the couples does not matter, we can use COMBINATIONS
We can select 3 couples from 5 couples in 5C3 ways ( = 10 ways)

ASIDE: If anyone is interested, we have a free video on calculating combinations (like 5C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

At this point, we have 3 COUPLES, which we'll call A, B ans C. We're now going to select ONE person from each couple to be on the committee.

Stage 2: Select 1 person from couple A
There are 2 people in this couple, so we can complete this stage in 2 ways.

Stage 3: Select 1 person from couple B
There are 2 people in this couple, so we can complete this stage in 2 ways.

Stage 4: Select 1 person from couple C
There are 2 people in this couple, so we can complete this stage in 2 ways.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create a 3-person committee) in (10)(2)(2)(2) ways ([spoiler]= 80ways[/spoiler])

Answer: D
--------------------------

Note: the FCP can be used to solve the majority of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by [email protected] » Fri May 08, 2015 3:00 pm
Hi prachi18oct,

There's nothing wrong with your approach; there's just a little more work that needs to be done...

With a group of 10 people, let's say that you select A, B and C. Since the question is asking for the number of different committees that can be formed, the order that the 3 people are selected DOES NOT MATTER.

This means that the following 6 groups are all the SAME group:

ABC
ACB
BAC
BCA
CAB
CBA

You're not allowed to count this group 6 times; you can only count it once.

Thus, the (10)(8)(6) = 480 groups that you calculated includes LOTS of duplicates. We have to divide this total by 6 to get the number of UNIQUE groups...

480/6 = 80

Final Answer: B

GMAT assassins aren't born, they're made,
Rich
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by prachi18oct » Fri May 08, 2015 3:09 pm
Thanks Rich! I completely missed that!
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by regor60 » Mon May 11, 2015 7:36 am
I did it by finding the number of ways to select 3 people from 10, which is 120.

Some of those groups of 3 include married couples, who never work well together as a general rule of married life.

There are 5 groups of 2 therefore who are disallowed, each paired then with 8 others to form the 3 person committee.

Disallowed 3 person committee then 5 x 8 =40. 120 - 40 = 80
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by [email protected] » Mon May 11, 2015 8:57 am
Hi regor60,

That approach works well too. You're going to find that most Quant questions can be approached in more than one way, so expanding your knowledge/tactics to include other approaches can be quite valuable.

In that same way, when reviewing CATs, you should look at ALL the questions (not just the ones that you got wrong) and look for ways that you could improve on the questions that you got CORRECT. Beyond practicing other tactics, maybe you could have organized your work better, taken better notes, spotted a Number Property (or other pattern), etc.

By having all of these various 'weapons' in your arsenal, Test Day will be all the easier.

GMAT assassins aren't born, they're made,
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