If a committee of 3 people is to be selected from among 5 married coup

In your selection of 5 people, say you selected 4 husbands and 1 wife (H1, H2, H3, H4, W5)
Now you selected 3 people out of these 5 say (H1, H2, H3)
This is one selection.

In your selection of 5 people, say you selected 3 husbands and 2 wives (H1, H2, H3, W4, W5)
Now you selected 3 people out of these 5 say (H1, H2, H3)
This is another selection.

But they both are the same selection but you are double counting them. That is why you are getting more than actual number.

I understood 3 approaches so far:
Approach #1:
Total # ways of forming committee of 3 members(without constraint) - # of ways in which 1 couple is present in the committee(constraint)
Total # ways of forming committee of 3 members(without constraint) = 10C3
# of ways in which 1 couple is present in the committee(constraint) = # of ways to select 1 couple out of 5 * # of ways of selecting 1 remaining individual out of 4 couples = 5C1 * 8C1

Approach #2:
# ways to select 3 couples out of 5 * # ways to select 1 person in C1 * ..C2 * ..C3 = 5C3*2C1*2C1*2C1

Approach #3:
We need to fill 3 seats in the committee
1st seat can be filled in 10C1 ways(selecting 1 out of 10 people)
2nd seat can be filled in 8C1 ways(selecting 1 out of 8 (=9 - spouse of 1st committee member))
3rd seat can be filled in 6C1 ways(selecting 1 out of 6 (=7 - spouse of 2nd committee member))
Total #of ways = 10C1*8C1*6C1
Now this answer includes order of selection as well, but we are not concerned about the order. # of ways to arrange 3 committee members is 3!. So dividing with the same to get the final answer

There are two distinct ways of approaching it:



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KarishmaB

When choosing 3 people from 5 couples, 10c1*8c1*6c1 = 10*8*6 = 480
When we use combination formula, we are already taking care that order doesnt matter. H1W2W3 is same as W2H1W3. Why do we divide by 3! after using combination formula when formula gives unique combination?

I do understand that when choosing 3 people from 10 using method like fill up 3 blanks
10 8 6
- - -
When we try to fill up this 3 blanks, now first position can be fill up 10 ways, then 8 and 6. In this way, we assume that order does matter and we need to divide by 3! to get unique combination. How it is different from 10c3?

i have different approach for this might be helpful:
so we have 5 couple and we have to form a group with unique member and constraints is community does not include two people who are married to each other.
i am building here there cases :
CASE 1: when all 3 member of community is male
>> there are 5 male in 5 couple so selecting 3 male out of 5 : 5c3 which is 10
CASE 2: when all 3 member of community is female
>>same goes for female as there are 5 female so selecting 3 from 5: 5c3 =10
now for 3rd case
CASE 3:
when 3 member from 5 couple can be with male or female and i am removing the couple after selected so that there is no repetition :
if we select 5C1*4C1*3C1 =60
Now add all the case: 60+10+10 = 80

choosing 3 couples out of 5 is 5C3=10. for each couple you have 2 choices : 2*2*2=8. 10*8=80

i have a doubt
5c3 is selecting non-couple people
and there are 4 possible cases
all men
all female
1 man 2 female
2 men 1 female

this concludes to 40 and not 80, what am i missing??

Hello ,
For selecting 3 people out of 5 married couple so that no married couple is selected
this can be done in following ways
total -committee with married couple
10c3-(5c1x8c1)
120-40=80

Hence option D is correct
I hope it helps

Thank you !!

there are 2 ways of solving the problem :

1.) 3 different people can be selected in 10C3 ways i.e. 120
now if we consider selecting a couple and one from the left our 8 people is 5C1 * 8C1 = 40
As it is not possible to select any other way the possibility of selection of 3 different people who are not married is 120-40= 80 ways

2.) filling spots could be done in 10C1*8C1*6C1 ways
And considering ABC=BCA=CAB... where order matters we can take only one in consideration
As 3 slots are to be filled that can be done on 3! ways
Therefore it could be done in 480/6= 80 ways

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

Total number of ways to select a committee of 3 out of 10 people without any restrictions = 10C3 = 120

Number of ways to select a committee of 3 with a couple = 5C1 * 8C1 =40

Number of committees without any couple = 120-40=80

IMO D