- #1

Dell

- 590

- 0

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