# Homework Help: Simple probability problem

1. Nov 18, 2009

### Dell

, what is the probability that no husband and wife are standing next to each other if:
there are 2 couples are standing in a line?
there are 3 couples are standing in a line?
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for 2 couples, my options are A,A',B,B' (where A and A' are a couple)

1st place (A/A'/B/B') => 4 options - say A is in 1st place...
2nd place (B/B') => 2 options - say B is in 2nd place...
3rd place (A') => 1 option
4th place (B') => 1 option

4*2*1*1=8 options that no husband and wife are standing next to one another.
4!=24 ways to place 4 people in a line

8/24=1/3
P(no couples)=1/3
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now for the 2nd case where there are 3 couples

1st place (A/A'/B/B'/C/C') => 6 options - say A is in 1st place...
2nd place (B/B'/C/C') => 4 options - say B is in 2nd place...
3rd place (A'/C/C') => 3 options - say C is in 3rd place...
4th place (A'/B') => 2 options - say A' is in 4th place...
5th place (B'/C') => 2 options - say B' is in 5th place...
6th place (C') => 1 option

6*4*3*2*2=288 options

<<OR>>

1st place (A/A'/B/B'/C/C') => 6 options - say A is in 1st place...
2nd place (B/B'/C/C') => 4 options - say B is in 2nd place...
3rd place (A'/C/C') => 3 options - say A' is in 3rd place...
4th place (C/C') => 2 options - say C is in 4th place...
5th place (B') => 1 option
6th place (C') => 1 option

6*4*3*2= 144 options

there are 6! ways to arrange the 6 people in the line

P(no couples)= (288+144)/6!=3/5

BUT THE CORRECT ANSWER IS ALSO MEANT TO BE 1/3

2. Nov 18, 2009

### willem2

Re: probability

Apperently your first case with 3 couples, is the case where the first 3 chosen persons are all from different couples. There will be only 2 persons to choose from in third place

Your second case with 3 couples is the case where the first 3 chosen persons are from only 2 different couples. There's only 1 person to choose from in third place