In how many ways to choose a group of 3 people from 6 couples so that

160

Number of ways 3 people can be chosen from 12 = 12C3 = 220

Number of ways a couple can be chosen = 6
Number of ways the third person can be chosen = 10C1 = 10

Hence total number of ways a couple with another person can be chosen = 6*10 = 60

Therefore total number of ways a couple cannot be chosen = 220-60 = 160

Just stumbled across this from a google search, so I'll add my technique.

first step: you can only choose one person from each couple, so I found how many combinations of couples I can have:

6c3=20

So there are 20 different ways to combine 3 of the 6 couples.

second step: once I have chosen the 3 couples to combine I have to choose man/wife from each. There are 2 choices in each of the 3 couples so I think of it as a binary number where 0=female and 1=male:

000 = woman,woman,woman
001 = woman,woman,man
...
111 = man,man,man

There are 8 combinations of man/woman from each of the 3 couples (2^3=8)

third step: multiply the number of couple combinations (6c3=20) times the number of man/woman combinations (2^3=8) == 160.

Voila!

Another single step calculation method is that you can select them using the simple basic counting principle (which arranges them in 1st person, 2nd person and 3rd person) and then you can divide by 3! to un-arrange.

First person can be selected in 12 ways.

Second person in 10 ways (since the first person selected and his/her partner are not available)

Third person in 8 ways (since the first and second people and their partners are not available)

Total number of ways = 12*10*8/3! = 160

Why do you divide it by 3!? what do you mean by un-arrange?

Thanks

When we use the method of placement of objects like 12*10*8 then we MUST take the cognizance of the arrangement of the objects already included in the method.
e.g. 5*4*3 = 5C3*3!
e.g. 10*9*8 = 10C3*3!


Similarly, 12*10*8 = Selected of three individuals such that no two form couple INCLUDING Arrangement of those three Individuals

therefore, we have to exclude the arrangements of those three individuals (3!) because the question only demand the no. of possible groups and not the arrangement of group members as well.

So only selection of those three individuals so that they are not couple = 12*10*8 / 3! = 160

I hope it helps!

Can somebody help me out with this?
Multiple doubts!

1. First approach was to select any of the 12 people. As there is another person who will correspond to being a couple there are 10 people and similarly 8 choices for the third place:
12*10*8

2. If I take the other approach and subtract no. of ways of selecting a couple and subtract it from the total ways to select 3, how do we get the number of selecting a couple as 60 as mentioned in some posts above?

1. First approach was to select any of the 12 people. As there is another person who will correspond to being a couple there are 10 people and similarly 8 choices for the third place:

12*10*8

This method includes the arrangement of those three individuals to be selected. Since we only require Selection so we shall exclude all arrangements of those three individuals by dividing the result by 3! (no. of ways in which 3 individuals can be arranged amongst each other)

i.e. 12*10*8/3! = 160

2. If I take the other approach and subtract no. of ways of selecting a couple and subtract it from the total ways to select 3, how do we get the number of selecting a couple as 60 as mentioned in some posts above?
Total ways to select 3 out of 12 people = 12C3 = 220
No. of ways selecting three people such that 2 of them are a couple = 6C1 * 10 = 60
6C1 = No. of ways of selecting one couple i.e. 2 people out of 6 couples
10 = No. of ways of selecting third person out of remaining 10 individuals (5 couples)
Favorable ways = 220 - 60 = 160

I hope this helps!

Thanks for the quick response. Could you please elaborate on this concept of dividing arrangements by factorial to get selections as I can't get my head around it. As per my opinion, I simply applied the counting method to get the answer.

If you have to select 3 individuals out of 5 people then you solve it like 5C3 = 10

But when we solve the question line filling three blank spaces by those 5 individuals then the method says

5 options for first place
4 options for Second place
3 options for Third place
i.e. total ways to fill three spaces with 3 of 5 people = 5*4*3 = 60
But this 60 includes the arrangement of those 3 individuals i.e. One of the arrangements of three people A, B and C 'ABC' will be treated different from arrangement 'ACB'

Now, This same question can be answered as 5C3*3! = 10*6 = 60

i.e. 5*4*3 = 5C3*3!
i.e. 5C3 = 5*4*3/3! Which is pure selection of 3 out of 5 individuals.

I hope this helps!

If a committee of 3 people is to be selected from among 6 married couples such 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) 80
(D) 120
(E) 160

Official Solution

Credit: Veritas Prep

We have 6 married couples i.e. we have 12 people. Out of these 12 people, we have to choose 3. If there were no constraints on the choice and we could choose any 3 out of these 12, in how many ways could we do that?

12 people and 3 available positions – We use basic counting principle to arrive at 12*11*10. But, recall that we have arranged the 3 people here. To un-arrange, we divide this by 3! to get 12*11*10/3! arrangements. This is what we discussed last week.

Moving forward, this question has a constraint – no two people married to each other should be selected i.e. at most one of any two married people could be selected. Say, if we select Mr. X, we cannot select Ms. X and vice versa.

Let’s try to use our basic counting principle and un-arranging method here:

You have 12 people and 3 positions. In the first position, you can put any one of the 12 people. In the second position, you can put any one of 10 people (not 11 because the spouse of the person put in 1st position cannot be put in the second position). In the third position you can put any one of 8 people. (We have already selected 2 people for the previous two places and their spouses cannot be selected for the third position so 4 of the 12 people are out.)

Total number of arrangements = 12*10*8

But mind you, these are arrangements. We just need a group of 3 people. They don’t need to be arranged in the group. So we divide these arrangements by 3! to just ‘select’ the people.

Number of committees possible = 12*10*8/3! = 160

Bunuel Could you please update the original Q with the answer choices? Thanks!

First select 3 couples out of the 6 possible.

6C3=6!/(3!*(6-3)!)=20

Then for each couple pick one of the persons to be on the board.

2^3=8

Multiply them together to get the final answer, E.

20*8=160