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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\frac { 4 } { 5 }

B

35\frac { 3 } { 5 }

C

25\frac { 2 } { 5 }

D

15\frac { 1 } { 5 }

Answer

45\frac { 4 } { 5 }

Explanation

Solution

Total ways =10!= 10 !

Two boys can sit side by side in 2×9!2 \times 9 ! ways.

So probaibility =2×9!10!=15= \frac { 2 \times 9 ! } { 10 ! } = \frac { 1 } { 5 }

Thus the probability that they are not seated together is 1−15=451 - \frac { 1 } { 5 } = \frac { 4 } { 5 }