Question

Three couples attend a dinner. Each of the six people chooses randomly a seat at a
round table. What is the probability that no couple sits together?​

Answer 1

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Let A1 be the event in which the first couple sit together.

Let A2 be the event in which the second couple sit together.

Let A3 be the event in which the third couple sit together.

So, the equation follows (6-1)!−A1−A2−A3+A1∩A2+A1∩A3+A2∩A3−A1∩A2∩A3

A1+A2+A3= 3 (5-1)! 2! =144

A1∩A2+A1∩A3+A2∩A3= 3 (4-1)! 2! 2! =72

A1∩A2∩A3 = (3-1)! 2! 2! 2! = 16

So,the no couple sits together is

120-144+72-16=32

The sample space is (6-1)!=120

The probabilty is

32/120 = 4/15