Problem Solving

gmatter2012 wrote:1.If a committee of 4 people is to be selected from among 6 married couples so that the committee does not include two people who are married to each other, how many such committees are possible? (None of them are married to each other)

gmatter2012 wrote:2. If a committee of 4 people is to be selected from among 6 married couples so that the committee includes exactly two people who are married to each other, how many such committees are possible? (One couple and other two non couple)

gmatter2012 wrote:3. If a committee of 4 people is to be selected from among 6 married couples so that the committee includes four people who are married to each other, how many such committees are possible? (Both couple)

Brent@GMATPrepNow wrote:

gmatter2012 wrote:1.If a committee of 4 people is to be selected from among 6 married couples so that the
committee does not include two people who are married to each other, how many such
committees are possible? (None of them are married to each other)

Here's one approach.


Aside: Notice that we must select 1 spouse from 4 of the 6 couples.

Take the task of selecting the 4 members and break it into stages.

Stage 1: Select the 4 couples from which we will select 1 spouse each.
There are 6 couples, and we must select 4. Since the order in which we select the 4 couples does not matter, this stage can be accomplished in 6C4 ways (15 ways)

Stage 2: Take one of the 4 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 4 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 4 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 5: Take one of the 4 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 5 stages (and thus create a 4-person committee) in (15)(2)(2)(2)(2) ways (= 240 ways)

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

Anurag@Gurome wrote: Number of ways of selecting 4 people from 6 couples (12 people) = 12C4 = 495
the number of ways to select 4 people who are non-couple = total number of ways - number of ways 1 couple and 2 non-couple - number of ways 2 couples
= 495 - 240 -15 = 240
But the solution Brent gave is the simpler one.

Anurag@Gurome wrote:
lets take the couples be (x1,y1)....(x6,y6)
In the numerator 12*1*10*1
lets take the 1st time you have chosen x1. so y1 will be chosen. the next time you have chosen x2. so y2 will be chosen.
what have you chosen is x1,y1,x2,y2 .........

Anurag@Gurome wrote:

so basically you are saying that 12*1*10*1 needs to be divided by an extra 2!
so what we have is this 12*1*10*1/2!2!2! = 15 am I correct ?

yes, you are right.

Suppose there are six couples , and six members need to be chosen ,Find ?

1)No. of ways exactly one couple can be selected
2) No. of ways exactly two couples can be selected
3)ways exactly 3 couples can be selected
4) ways such all are singles , i.e no couple is chosen?


one couple
6C1 *5C4 *(2)^4= 480

two couple
6C2*4C2*(2)^2= 360

Three couple
6C3= 20

No couple
6C6 * (2)^6= 64

** ** ** *** **


Suppose there are Five couples and we need to select 4, how to find?

1)No. of ways exactly one couple can be selected
2) No. of ways exactly two couples can be selected
3) ways such all are singles , i.e no couple is chosen?

One couple only

5C1 *4C2*(2)^2= 120

two couple only

5C2= 10


All singles

5C4 *(2)4 = 5 *16= 80

Hope these are correct?

The answers look good.