One way to approach:
Desired No. of Committees = Total Committees Possible - Committees you don't want to form
Total no. of people = 10; No. of ways to select 3 out of them = 10C3 = 120
This includes scenarios when couples are also together.
Now, if we include a couple then the 3rd person can be selected from the remaining 8 people, in
8C1 ways. There are 5 possible couples, so total = 5 * 8C1 = 40
Hence, total no. of committees, without couples together = 120-40 = 80 (Option D)

Alternate method:
Each couple can send only one "representative" to the committee. Let's see in how many ways
we can choose 3 couples (as there should be 3 members) to send only one "representatives" to
the committee: 5C3 = 10.
But these 3 couples can send two persons (husband or wife): 2*2*2=23=8.
Total # of ways: 5C3 * 23 = 80 (Option D)

Another approach:

We need to choose a committee of 3 people:

1st person: choose any of the 10

2nd person: choose any of the remaining 8 (since we can't have the husband/wife of the first person)

3rd person: choose any of the remaining 6

We can order 3 people: 3! = 6 different ways

10*8*6 / 6 = 480/6 = 80

Answer is D.
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Total number of ways in which 3 people can be chosen out of 10 is equal to 10X9X8 DIVIDED by 3X2X1 = 120.

Total number of ways in which 3 people can be chosen given that two MUST be couples can be represented as M-M-N in which M = married and N = not married.

First M:
So for the first M there are 10 different people we can choose.

Second M:
The second M must be equal to 1 since only 1 person can be married to the first M selected.

The N:
The N can be selected from 8 different people after removing the couple from the group.

So the MMN is equal to 10X1X8 and we must divide by 2 to account for the two "duplicate" M's. So 120 - 40 = 80 represents the number of combinations in which there is no married couples selected.

We have 5 couples, and 10 people total.


For any given committee of 3 people that doesn’t have a married couple on it, 1 partner from a couple will be “donated” from that couple to the 3 person committee.


(1st) in how many ways can we choose 3 couples that will “donate” 1 partner to the committee if there are 5 different couples?

5 c 3 = 5! / 3!2! = 5 * 4 / 2 = 10 ways

AND

(2nd)for each of these 10 ways to have 3 couples chosen, we have to pick 1 person from each of the 3 couples to be on the committee:


Couple 1: 2 available options

Couple 2: 2 available options

Couple 3: 2 available options


(10) * 2 * 2 * 2 =

80 ways

Method 2:

Pick the 1st person for the committee: 10 available options

And

Picks 2nd person: we can not pick that prior person’s spouse —- so 8 available options

And

Pick 3rd person: we can not pick the spouses of the prior 2 people selected already —- so 6 available options


10 * 8 * 6

However, this will include the number of ways to select 3 people as well as arranging those 3 people within each possible combination. For such 1 unique combination, we have counted (3!) —— thus divide by (3!) to get rid of the “arrangements” within each possible combination/group of 3 people


(10 * 8 * 6) / (3!) =

80 ways

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VeritasKarishma any idea why my approach didnt work


total number of ways: 5C3 = 120

no i choose a couple: so there are 10 ways to pick any person and 1 way to pick a match and 8 ways to pick any third person

so 10*1*8 = 80

120-80 =40

You can pick 3 people out of 10 in 10C3 = 120 ways. No problem.

Now, we are not allowed to pick a couple among these 3 people. So cases in which we have a couple are not allowed.
How many such cases are there? Pick one couple out of 5 couples in 5C1 = 5 ways. Pick the third person in 8 ways. So combinations with a couple = 5*8 = 40

Acceptable combinations = 120 - 40 = 80


dave13 : Note that you cannot pick a couple in 10 ways. You are double counting each couple then. Say you pick HubbyA and then pick WifeA is one case. If you pick WifeA and then HubbyA, you are counting this as another case but it is not.
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VeritasKarishma thanks so if i am double counting using this method 10*1*8 = 80 then i divide by 2! and get 40. is my method correct now ?

Yes, but picking 10 and then dividing by 2! should make logical sense to you.
I know there are 5 couples so I can pick one couple in 5 ways. That is what makes sense to me.
I am finding it hard to explain to myself why I would pick 10 and then divide by 2!.
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