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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Thu Dec 10, 2009 2:53 pm

valleeny wrote:Hi

Can someone explain to me what is wrong with my method.

To get 3 person without a married couple in it,

1) First, select any 3 couples out of 4 from which the committee will be selected. There are 4C3 ways to do it.
2) Each couple can only contribute one member. Therefore, there are 2C1 * 3 ways to do it.

Ans 4C3 * 2C1 * 2C1 * 2C1 = 4C3 * 2C1 * 3 = 24.

What is wrong?

Excellent reasoning... bad math!

2C1 * 2C1 * 2C1 does not equal 2C1 * 3, it equals 2C1 ^ 3

So:

4C3 * 2C1 * 2C1 * 2C1 = 4 * 2 * 2 * 2 = 32... correct answer.

(I stared at your post for a long time, trying to find the flaw in your reasoning and I couldn't - then I actually checked the arithmetic.)

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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valleeny: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Mon Nov 16, 2009 4:54 pmFollowed by: 1 member; Thanked: 8 times

by valleeny » Thu Dec 10, 2009 4:05 pm

Thank you Stuart. That was bad on my part. Can't believe I didn't spot that even after I kept repeating the steps. It's cleared now. Thank you.

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valleeny: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Mon Nov 16, 2009 4:54 pmFollowed by: 1 member; Thanked: 8 times

by valleeny » Thu Dec 10, 2009 4:09 pm

Thank you Stuart. That was bad on my part. Can't believe I didn't spot that even after I kept repeating the steps. It's cleared now. Thank you.

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chandni170: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Fri Jul 05, 2013 1:07 am

by chandni170 » Sun Jul 07, 2013 1:05 am

Hi,

I remember doing this type of a sum long back but I'm kind of stuck. Not sure what I'm supposed to do after the second step.

Here's my solution:

Let the couples be: AB CD EF GH

To ensure the committee doesn't have a married couple, I separate the couples in my first pick:

2C1*2C1*2C1*2C1= 16 ways to select 4 people (this ensures that I will not have a couple in my committee)

Now I can pick 3 out of the 4 chosen people in 4C3 ways= 4 ways.

Therefore, total ways to form the committee = 16x4= 64ways... which is wrong

Also, I usually get confused when to add the number of ways (that I've calculated separately) and when to multiply them. For example, once I've calculate the 16 ways and 4 ways above, I'm unsure whether I should multiply them or add them.

Experts, please could you clarify my approach here?

Thanks

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sun Jul 07, 2013 5:16 am

dhanda.arun wrote:A committee of 3 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 cannot both serve on the committee?
A) 16
B) 24
C) 26
D) 30
E) 32

Take the task of selecting the 3 committee members and break it into stages.

Stage 1: Select the 3 couples from which we will select 1 spouse each.
There are 4 couples, and we must select 3 of them. Since the order in which we select the 3 couples does not matter, this stage can be accomplished in 4C3 ways (4 ways)

If anyone is interested, we have a free video on calculating combinations (like 4C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 3: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 4: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus create a 3-person committee) in (4)(2)(2)(2) ways (= 32 ways)

Answer = E

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

Brent Hanneson - Creator of GMATPrepNow.com

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