Combinations 3

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.

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?

I am not pretty sure but i think it should be 80.

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,

goyalsau wrote:

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?

I am not pretty sure but i think it should be 160.

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,

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.

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.

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".

Can you please explain.

The Jock wrote:

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.

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".

Can you please explain.

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.

Thanks a lot. got it now

goyalsau wrote:

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?

I am not pretty sure but i think it should be 160.

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,

logically we can also see that 10C3=120 ,so the answer should be less than 120.

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.

Thanks Sir, I have edited it, I did it correctly but marked the answer wrongly,

I have some difficulty understanding TestLuv explanation.


I divided the set into H1,H2...H5 and W1,W2...W5...
H - Husband, W- Wife and the number indicates the couple number.

So if we need a community of people who dont include those in the couples together,Then our selection should be

1. From the husbands group
H1, H2, H3, H4, H5

Select 3 out of the 5 individuals.

5C3 ways = 10 ways

2. From the wife's group

W1,W2,W3,W4, W5

Select 3 out of 5 individuals.

5C3 ways = 10 WAys

3. 2 From Husband group and 1 from wife group(Excluding the one's apart from the husband group).
H1 H2 H3 H4 H5
W1 W2 W3 W4 W5

Here if our selection is H1 and H2 then Other person can be W3,W4,W5.

So essentially i thought this comes down to 5C2 For the husband Group 10 ways.

and Remaining wife groups are W3,W4,W5 out of which we have to select one more person

so its 3C1 ways which is 3 ways.

( So multiplying 10 x 3 = 30 WAys)

4. 2 From Wife group and 1 from Husband group(Same condition as before)
5C2 ways for Wife group
3C1 Ways for Husband Group

So multiplying again 10 x 3 = 30 ways

(30+30+10+10) = 80 Ways...

Any other shortcut other than this ???