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(B) Total number of ways to arrange $6$ students in a row $= 6! = 720$.To find the number of favorable arrangements, consider $A$ and $B$ with one stu

Vedclass · Accepted Answer

$\frac{4}{15}$

Vedclass · Answer

(B) Total number of ways to arrange $6$ students in a row $= 6! = 720$.To find the number of favorable arrangements, consider $A$ and $B$ with one student between them as a single block. There are $4$ remaining students. We choose $1$ student out of $4$ to place between $A$ and $B$ in $^4C_1$ ways.Now, treat the block $(A, 	ext{student}, B)$ as one unit. Along with the remaining $3$ students, we have $4$ units to arrange, which can be done in $4!$ ways.Within the block, $A$ and $B$ can be arranged in $2!$ ways (i.e.,$A, 	ext{student}, B$ or $B, 	ext{student}, A$).Number of favorable arrangements $= ^4C_1 	imes 4! 	imes 2! = 4 	imes 24 	imes 2 = 192$.Probability $= \frac{192}{720} = \frac{4}{15}$.

If a group of six students including two particular students $A$ and $B$ stand in a row, then the probability of getting an arrangement in which $A$ and $B$ are separated by exactly one student in between them is

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