I am not pretty sure but i think it should be 80.joannabanana wrote:A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?
Hi,goyalsau wrote:I am not pretty sure but i think it should be 160.joannabanana wrote:A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?
There are in all 10 people with 5 couples,
so 1st place can be filled in 10 ways,
2nd place in 8 ways,
3rd place in 6 ways,
10 * 8 * 6 = 480
480/6 = 80 community,
Divided by 6 Because I think with the above combination it will form 6 committee with the same group,
LIKE
A1, B1, C1, Can be arranged in 2 ways, A1, C1, B1,
B1,A1, C1, Can be arranged in 2 ways B1, C1, A1
C1,B1,A1, Can be arranged in 2 ways C1, A1, B1
I don't know whether i am doing it correctly But lets see, What's out Quant Guru's have to say on this,
Finding a little difficulty in underatanding following line "The restricted selections can be computed by selecting any of the 5 couples, and then any person from the remaining 8 individuals or 5*8".Testluv wrote:This is a combination problem involving a restriction, so recall that:
total number of selections = restricted selections + permitted selections
or
permitted selections = total number of selections - restricted selections
The total number of selections is just all the ways we can pull 3 people out of 10 or 10C3. The restricted selections can be computed by selecting any of the 5 couples, and then any person from the remaining 8 individuals or 5*8.
So we have:
permitted selections = 10C3 - (5*8) = 80.
Sure thing. To begin, we can select any 1 of the 5 couples or 5C1. The third person can be any of the remaining 8 individuals (after we select a couple, there's 8 people left) or 8C1. So, the restricted selections are 5C1*8C1 or, more simply, 5*8.The Jock wrote:Finding a little difficulty in underatanding following line "The restricted selections can be computed by selecting any of the 5 couples, and then any person from the remaining 8 individuals or 5*8".Testluv wrote:This is a combination problem involving a restriction, so recall that:
total number of selections = restricted selections + permitted selections
or
permitted selections = total number of selections - restricted selections
The total number of selections is just all the ways we can pull 3 people out of 10 or 10C3. The restricted selections can be computed by selecting any of the 5 couples, and then any person from the remaining 8 individuals or 5*8.
So we have:
permitted selections = 10C3 - (5*8) = 80.
Can you please explain.
logically we can also see that 10C3=120 ,so the answer should be less than 120.goyalsau wrote:I am not pretty sure but i think it should be 160.joannabanana wrote:A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?
There are in all 10 people with 5 couples,
so 1st place can be filled in 10 ways,
2nd place in 8 ways,
3rd place in 6 ways,
10 * 8 * 6 = 480
480/6 = 80 community,
Divided by 6 Because I think with the above combination it will form 6 committee with the same group,
LIKE
A1, B1, C1, Can be arranged in 2 ways, A1, C1, B1,
B1,A1, C1, Can be arranged in 2 ways B1, C1, A1
C1,B1,A1, Can be arranged in 2 ways C1, A1, B1
I don't know whether i am doing it correctly But lets see, What's out Quant Guru's have to say on this,
Thanks Sir, I have edited it, I did it correctly but marked the answer wrongly,Testluv wrote: Hi,
your solution is sound! But the answer is 80 (not 160). You were right to divide by 6 because there are 3! ways of arranging three people. Another approach is outlined in my first post, above.
