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computation problem [#permalink]
24 Aug 2004, 18:54

Please provide your time in solving this problem along with your answer:

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I think I have seen this type of question before.
Say 8 persons are A,B,C,D,E,F,G,H.. AB,CD,EF,GH are couples
Calculate by reducing number of ways when couples will be together from number of total ways.
Number of total ways..8C3
Number of ways when couples will be together.. Say A, B are together..in this case number of ways to select 3rd person from remaining 6 is 6C1=6
Similarly when C,D are together..number of ways to selct 3rd person is 6.
When E, F are together..another 6 ways.
G,H are together..another 6 ways..
So total 24 ways.. when couples are together.
Hence required number of ways = 8C3 -24
=56-24
= 32
So 32 should be the answer

Consider the couples as (H1,W1), (H2,W2), (H3,W3), (H4,W4), (H5,W5)
For each pair of couple they can pair up with either the husband or wife of another pair to form a group of 3
So for example, H1,W1,H2, H1,W1,W2..etc. So one couple can have total of 8 possible pairings
5 couples will give 5*8=40 pairs

So total number of 3 people group with husband and wife not in the saem group = 120-40 = 80

Re: computation problem [#permalink]
24 Aug 2004, 22:46

I think it is 80.

10 members. So, 10C3 = 120 is the number of groups formed. This 120 may contain even the couples.

Groups that contain couples only = 5C1 * 8C1 = 40

So, the groups that donot contain couples are 120-40 = 80

lastochka wrote:

Please provide your time in solving this problem along with your answer:

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

1) Total - unwanted.
10C3 - 5*8 = 80
5 is for choosing one married couple of the 5, and 8 is for choosing another person out of the remaining8 .

2) 10*8*6/3! = 80.
For the first person you have 10 options.
For the second, only 8.
For the third, only 6.
Divide by 3! because the order doesn't matter.

mallelac, do you mind explaining how you came to the formula for the undesireable outcomes? Wouldn't 5C1+8C1 only give you the number of people that would comprise a 2 person team?

The undesirable outcome should have a couple. We can choose one out of 5 couples in 5C1 ways thus filling two spots(not one spot). Now, 10-2 = 8 members can be fill up the third spot in 8C1 ways. Thus total ways of undesired outcomes = 5C1*8C1 = 40.

Do let me know if I am not clear.

abennett wrote:

mallelac, do you mind explaining how you came to the formula for the undesireable outcomes? Wouldn't 5C1+8C1 only give you the number of people that would comprise a 2 person team?