Problem Solving

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?
16
24
26
30
32


OA is 32

8 total people and 3 are to be chosen.

8C3

8!/3!(8-3)!
8!/3!5!
56 Total ways

Now you know that the set can't include married couples.


To find the number of ways married couples could be part of a set you do the follwoing:

4 Married couples multiplied by the additional 6 people = 24 ways to have married couples together.



56 total ways - 24 ways with married couples = 32 ways without married couples

First person can be selected in 8 ways [can be any one]
second person can be selected in 6 ways [keep aside the spouse of person just selected]
thrd person can be selected in 4 ways [keep aside the spouse of two persons selected]

total 8 * 6 * 4

now our selection is order dependent
so we need to divide it by 3! to make it indep of order
Remember ABC and BAC are same committee

has this been some prize and we had first second and third prize
than order is imp
so we do not need to divide by 3!

8 * 6 * 4 / 3! = 32

There is another method also using combinations
but I refrained from that as it may confuse you

HI,
Thanks for the solution
Please share the other method also.

dhanda.arun wrote:HI,
Thanks for the solution
Please share the other method also.

Already given by dmateer25.

dmateer25 wrote:8 total people and 3 are to be chosen.

8C3

8!/3!(8-3)!
8!/3!5!
56 Total ways

Now you know that the set can't include married couples.


To find the number of ways married couples could be part of a set you do the follwoing:

4 Married couples multiplied by the additional 6 people = 24 ways to have married couples together.



56 total ways - 24 ways with married couples = 32 ways without married couples

Can you please explain me this part ... ??

4 Married couples multiplied by the additional 6 people = 24 ways to have married couples together.

if one couple takes the first two spots: we have 6 people competing for the last spot.

So 4 couples x 6 = 24 ways ( Couples can be in the same group)

8C3= 56

56 - 24 = Correct Answer

Hi,

Can anyone explain how did we arrive at 3 !. I understood the first part.

Regards,
Shobha

stop@800 wrote:First person can be selected in 8 ways [can be any one]
second person can be selected in 6 ways [keep aside the spouse of person just selected]
thrd person can be selected in 4 ways [keep aside the spouse of two persons selected]

total 8 * 6 * 4

now our selection is order dependent
so we need to divide it by 3! to make it indep of order
Remember ABC and BAC are same committee

has this been some prize and we had first second and third prize
than order is imp
so we do not need to divide by 3!

8 * 6 * 4 / 3! = 32

There is another method also using combinations
but I refrained from that as it may confuse you

getso wrote:Hi,

Can anyone explain how did we arrive at 3 !. I understood the first part.

Regards,
Shobha

stop@800 wrote:First person can be selected in 8 ways [can be any one]
second person can be selected in 6 ways [keep aside the spouse of person just selected]
thrd person can be selected in 4 ways [keep aside the spouse of two persons selected]

total 8 * 6 * 4

now our selection is order dependent
so we need to divide it by 3! to make it indep of order
Remember ABC and BAC are same committee

has this been some prize and we had first second and third prize
than order is imp
so we do not need to divide by 3!

8 * 6 * 4 / 3! = 32

There is another method also using combinations
but I refrained from that as it may confuse you

Let's say you had a pair of objects: A and B. If order doesn't matter, then how many pairs do you have? Clearly AB is one pair. And if order doesn't matter, then it doesn't matter whether you call that pair AB or BA--it's still one pair--or one "unordered" group of two objects. But if order matters, then AB is one "ordered" pair and BA is another ordered pair. So, notice that when order matters we have 2 pairs, and when order doesn't matter it's just 1 pair. To go from the number of ordered groups to the number of unordered groups, you have to divide by the factorial of the number of objects. A and B are two objects. There are two ordered pairs. To remove order: 2/2! = 1 unordered pair.

Let's say you have three objects: A, B and C. That's one unorderered group of three. But there are 3! or six ordered groups (number of ways of arranging n objects is n!). To go from from the number of ordered groups to the number of unordered groups, we would just divide the number of ordered groups--6--by the factorial of the number of objects--3!, giving us 1.

Now, in the above two paragraphs I was talking about groups. But this problem and GMAT combinatorics frequently involves selecting subgroups from a larger group. The selection of unordered subgroups are combinations; the selection of ordered subgroups are permutations. Let's consider the formula for selecting combinations and permutations from bigger sets. Let's first look at the permutations formula:

nPk = n!/(n -k)!

in which n is the number of objects in the big group and k is the number of objects in the subgroup. This formula yields the total number of all the ordered subgroups of size k selected from a bigger group of size n

Now, let's look at the formula for combinations:

nCk = n!/[k!*(n-k)!]

This formula yields the total number of all the unordered subgroups of size k selected from a bigger group of size n. So, notice that the only difference in the two formulae is that in the combinations formula we are, again, dividing by the factorial of the number of objects in the subgroup. So, the only difference between the two formulae acknowledges that when we are unconcerend about order, we must divide by the factorial of the number of objects in the group or subgroup.

In this problem, when we place any of the eight people in the first slot, leaving six for the second, and four for the third, ABC will end up being counted as a different committee from BCA. But because they are comprised of the same people, ABC is the same committee as BAC, no matter whether you call them "Andy, Bob, and Charlie" or "Bob, Andy, and Charlie". In other words, in this problem, order does not matter. This is why we have to divide 8*6*4 by the factorial of the number of objects in the subgroup (committee)--3!.

In combinatorics problems, you should always determine very early on whether order matters or not. In order for order to matter, the problem has to communicate a situation where order would matter: for example, ranking or a code, or using the word "order" or "arrange", etc.

Thanks Testluv for the great explaination.

Yes I get it now ....why we should divide by factorial of a number.

Regards,
Shobha

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?

Firstly, the question says that 3 people must be selected and NOT 3 couples. You are actually trying to select 3 couples (6 people) out of 4 couples (8 people).

Second one, i'm not sure of what you are doing.

I know the question asks for 3 persons and not couple.

My train of thought is that if I am going to select these 3 members, the process will be as follows.

1) Single out 3 couples out of 4 (selected 6 people out of 8)
2) From each of the 3 singled out couples, pick just 1 to become a member (selected 3 people out of 6)
3) In the end, you end up with 3 non married members from 3 different couples

Very confusing train of thought but can you point out the flaw in this thought process?