Hi all,
I'm terrible with combinations...there is a question here for which I don't agree with the solution: Could anyone help elucidate?
1. Two couples and one single person sit in random in a row of 6 chairs. What it the probability that neither of these couples sit together in adjacent chairs?
I would have:
Find Probability that the couples are "one unit". So
Each couple=1 unit,and single person=1 unit
So there are 3 "units" total
6C3 --> 6!/3! = 6 x 5 x 4 x 2 x 2 = 480<---multiply by 2*2 because within each couple, the people can switch.
Probability of ONE couple sitting together and the other not
C1 A1 A2 B C2
C2 A1 A2 B C1
A1 A2 C1 B C2
A1 A2 C2 B C1
--> 4 x 2 x 2 (multiply by 2 because of the exact opposite orientation, and by 2 again because it can be C1 C2 that sit together instead)
= 16
Total possibilities = 6C5 (now treat couples as separate people) = 6 x 5 x 4 x 3 x 2= 720
So probability of sit together = (480 + 16)/720 = 496/720. Probability of NOT sit together is 1-496/720
This is clearly not the answer...can someone tell me why?
In short, this is the answer below:
I sort of understand it but don't agree with the line I marked in asterisk. Why would you divide by 2 if you want to count the two couples as the same? Aren't they different units??
https://www.beatthegmat.com/download-gma ... 59366.html
Solution:
Let [] ==> denotes each chair
Let the couple be denoted by C1 and C2 and the other single person By S
Please draw 6 chairs in some scratch paper for better understanding
[] [] [] [] [] []
Lets first sit C1
[C1] [] [] [] [] []
Now C2 can take only the adjacent seat
[C1] [C2] [] [] [] []
now S can seat in 4 different ways(4 empty chairs!)
Although, C1 and C2 can also be interchanged as C1 C2 or C2 C1 which I will deal latter in this question
Similarly,
for another Seating arrangement like this
[] [C1] [C2] [] [] []
Now again S can be seated in 4 different ways, one on left of C1 and 3 on right of C2
[] [] [C1] [C2] [] []
Again 4 types of seating arrangements for S
Now if we further advance C1 C2 to the right, the Second arrangement again occurs
[] [] [] [C1] [C2] []
So we are not going to count this arrangement.
So at Max, there can be 3 different arrangements(Bold Face) having 4 types of seating arrangement of S
thus, 3*4 = 12
Further, this seating arrangement of C1 C2 Can be interchanged as C2 and C1
12*2 = 24
And Total Number of Arrangements of 3 people in 6 chairs is
6*5*4/2! =120/2 = 60
***************two people as couples (same type) = 2! ***************
So ways of couple not seating together is 60-24 =36
So probability = 36/60 =3/5
I'm terrible with combinations...there is a question here for which I don't agree with the solution: Could anyone help elucidate?
1. Two couples and one single person sit in random in a row of 6 chairs. What it the probability that neither of these couples sit together in adjacent chairs?
I would have:
Find Probability that the couples are "one unit". So
Each couple=1 unit,and single person=1 unit
So there are 3 "units" total
6C3 --> 6!/3! = 6 x 5 x 4 x 2 x 2 = 480<---multiply by 2*2 because within each couple, the people can switch.
Probability of ONE couple sitting together and the other not
C1 A1 A2 B C2
C2 A1 A2 B C1
A1 A2 C1 B C2
A1 A2 C2 B C1
--> 4 x 2 x 2 (multiply by 2 because of the exact opposite orientation, and by 2 again because it can be C1 C2 that sit together instead)
= 16
Total possibilities = 6C5 (now treat couples as separate people) = 6 x 5 x 4 x 3 x 2= 720
So probability of sit together = (480 + 16)/720 = 496/720. Probability of NOT sit together is 1-496/720
This is clearly not the answer...can someone tell me why?
In short, this is the answer below:
I sort of understand it but don't agree with the line I marked in asterisk. Why would you divide by 2 if you want to count the two couples as the same? Aren't they different units??
https://www.beatthegmat.com/download-gma ... 59366.html
Solution:
Let [] ==> denotes each chair
Let the couple be denoted by C1 and C2 and the other single person By S
Please draw 6 chairs in some scratch paper for better understanding
[] [] [] [] [] []
Lets first sit C1
[C1] [] [] [] [] []
Now C2 can take only the adjacent seat
[C1] [C2] [] [] [] []
now S can seat in 4 different ways(4 empty chairs!)
Although, C1 and C2 can also be interchanged as C1 C2 or C2 C1 which I will deal latter in this question
Similarly,
for another Seating arrangement like this
[] [C1] [C2] [] [] []
Now again S can be seated in 4 different ways, one on left of C1 and 3 on right of C2
[] [] [C1] [C2] [] []
Again 4 types of seating arrangements for S
Now if we further advance C1 C2 to the right, the Second arrangement again occurs
[] [] [] [C1] [C2] []
So we are not going to count this arrangement.
So at Max, there can be 3 different arrangements(Bold Face) having 4 types of seating arrangement of S
thus, 3*4 = 12
Further, this seating arrangement of C1 C2 Can be interchanged as C2 and C1
12*2 = 24
And Total Number of Arrangements of 3 people in 6 chairs is
6*5*4/2! =120/2 = 60
***************two people as couples (same type) = 2! ***************
So ways of couple not seating together is 60-24 =36
So probability = 36/60 =3/5
