3 couples seated randomly around a round table. What's the chance 1 or more of the couples will be together?

3 couples. They are randomly seated around a round table. What is the probability that any one or more of the couples will be seated together -- husband to wife's left or right, doesn't matter.3 couples seated randomly around a round table. What's the chance 1 or more of the couples will be together?
First, there are (n-1)! ways to arrange n people in a circle. So there are (6-1)!=120 ways to sit these people.


Find probability of no couples together, then subtract 1 from it.


Fix a person. His/her mate can sit either one seat to left or one seat to right. Take left. The remaining couples can be organized in (2x2x2) ways. This is also true for the right. (2x) So there are 2x2x2x2=16 ways for now. His/her mate can also sit exactly in front. There are 2x2x2x2=16 ways the remaining couples can sit. In total there are 32 ways. 32/120=.27. So the probability of at least one couple together is .73. This solution agrees with simulation results.





EXTRA:


Probability of exactly one couple.


Fix a couple seated together(2x). The other couples can only sit one seat apart. They can exchange places. (2x2) They can exchange seats with the other couple (2x). There are 2x2x2x2=16 positions when one couple is fixed. Since fixing others will form completely different seatings, the total number of possible seatings with one couple is 16*3=48. The probability is 48/120=.4.





Probability of exactly two couples.


Fix a couple A (2x) and combine the other couple say B together(2x). B can only sit in front of A, since if they sit next to each other there will be 3 couples sitting together. The remaining couple can change places (2x). We can select A and B in three possibilities (3x). So the number of possible seatings is 2x2x2x3=24. The probability is 24/120=.2.





Probability of exactly three couples. The three couples sit together. Any of them can change seats (2^3x). Only (3-1=2) ways to organize 3 items in a circle. So, the possible seatings is (2^3x2=16). The probability is 16/120=.13.





The total number of seatings is 32+48+24+16=120.3 couples seated randomly around a round table. What's the chance 1 or more of the couples will be together?
at least one couple is together: fix that couple... there will be 4 people to arrange... the couple can change places


number of ways = 4! x 2 = 48











at least two couples are together: fix one couple...


there are 3 entities to arrange...the couples can change places


number of ways = 3! x 2 x 2 = 24





all 3 couples are together


number of ways = 2x2x2x2 = 16








number of ways that at least one couple is seated together = 48 - 24 + 16 = 40








total ways of arranging 6 people in a circle... = 5! = 120








thus the total probability = 40/120


= 1/3











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