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Question: Ten students are seated at random in a row. The probability that two particular students are not sea...

Ten students are seated at random in a row. The probability that two particular students are not seated side by side is?
(a) 45\dfrac{4}{5}
(b) 35\dfrac{3}{5}
(c) 25\dfrac{2}{5}
(d) 15\dfrac{1}{5}

Explanation

Solution

First find the number of possible ways to arrange ten students in the row without any condition (total number of arrangements). Now, consider the situation that the two particular students are always seated together and find the possible ways of arrangement in that case by considering those two students as a single unit. Subtract the number of arrangements obtained in two different cases to get the number of favourable arrangements. Divide the number of favourable arrangements by the total number of arrangements to get the answer.

Complete step-by-step solution:
Here there are ten students who are to be seated in a row with the condition that two particular students will not sit next to each other. According to the question it is already fixed that two students must not be seated side by side so we are not going to select them particularly.
Now, consider the case that the ten students are to be seated without any conditions, so we need to arrange 10 students at 10 places. So we have,
⇒\Rightarrow Number of arrangements possible = 10!10!
Now, let us consider the case where we assume that the two particular students are always seated side by side, so we can consider them as one unit or roughly we can consider them as a single person. Therefore, now we have 9 students to arrange.
⇒\Rightarrow Number of arrangements possible = 9!9!
Also, these two particular students can be arranged among themselves in 2!2! ways. So we have an effective number of arrangements =9!×2!=9!\times 2!.
Therefore, the total number of arrangements where these two particular students are not seated side by side =10!−(9!×2!)=10!-\left( 9!\times 2! \right).
So, the total number of arrangements is 10!10! and the number of favourable arrangements is 10!−(9!×2!)10!-\left( 9!\times 2! \right). The probability will be the ratio of number of favourable arrangements to the total number of arrangements, so we get,
⇒\Rightarrow Required probability = 10!−(9!×2!)10!\dfrac{10!-\left( 9!\times 2! \right)}{10!}
⇒\Rightarrow Required probability = 1−9!×2!10!1-\dfrac{9!\times 2!}{10!}
⇒\Rightarrow Required probability = 1−151-\dfrac{1}{5}
∴\therefore Required probability = 45\dfrac{4}{5}
Hence, option (a) is the correct answer.

Note: Note that if we will try to find the favourable arrangements directly then it will become a bit complicated as there are 10 students (10 seats) and to form arrangements according to the condition given is difficult. This is the reason we have followed the above process. Also, note that we do not have to select the two particular students in 10C2^{10}{{C}_{2}} ways because they are already chosen and fixed in the question.