0
$\begingroup$

Two couples and a single person are seated at random in a row of five chairs. What is the probability at least one person is not beside his/her partner?

Let $P(A)$ denote the probability that both couples are seated together in the arrangement. Therefore, the probability that at least one person is not beside his/her partner should be
$$1-P(A)$$ The total number of arrangements is $5!$ and the number of arrangements for both couples sitting together is $$3!2!2!$$ So, the probability that at least one per son is not beside his/her partner is $$ 1-\frac{3!2!2!}{5!}=\frac{4}{5}$$

Are there any errors in the solution and answer?

$\endgroup$
0
$\begingroup$

No there is no errors, your solution is right

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.