Ten students are seated at random in a row. T | vedclass

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(A) Total ways to arrange $10$ students in a row is $10!$. To find the probability that two particular students sit together,we treat them as a single

Vedclass · Accepted Answer

$\frac{4}{5}$

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(A) Total ways to arrange $10$ students in a row is $10!$. To find the probability that two particular students sit together,we treat them as a single unit. There are $9$ units (the pair + $8$ other students),which can be arranged in $9!$ ways. The two students within the pair can be arranged in $2! = 2$ ways. So,the number of ways they sit together is $2 	imes 9!$. The probability that they sit together is $\frac{2 	imes 9!}{10!} = \frac{2 	imes 9!}{10 	imes 9!} = \frac{2}{10} = \frac{1}{5}$. The probability that they are not seated side by side is $1 - \frac{1}{5} = \frac{4}{5}$.

Ten students are seated at random in a row. The probability that two particular students are not seated side by side is

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