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(B) Total number of ways to arrange $6$ students in a row is $6! = 720.$To find the number of arrangements where $A$ and $B$ are separated by exactly

Vedclass · Accepted Answer

$\frac{4}{15}$

Vedclass · Answer

(B) Total number of ways to arrange $6$ students in a row is $6! = 720.$To find the number of arrangements where $A$ and $B$ are separated by exactly one student,we treat the block $(A, X, B)$ or $(B, X, A)$ as a single unit,where $X$ is one of the remaining $4$ students.Step $1$: Choose one student $X$ from the remaining $4$ students in $4$ ways.Step $2$: Arrange the block $(A, X, B)$ or $(B, X, A)$ along with the remaining $3$ students. This gives us $4$ units to arrange,which can be done in $4!$ ways.Step $3$: Since the block can be $(A, X, B)$ or $(B, X, A)$,there are $2$ ways to arrange $A$ and $B$ within the block.Total favorable arrangements $= 4 	imes 4! 	imes 2 = 4 	imes 24 	imes 2 = 192.$Probability $= \frac{192}{720} = \frac{192 \div 48}{720 \div 48} = \frac{4}{15}.$

If six students,including two particular students $A$ and $B,$ stand in a row,then the probability that $A$ and $B$ are separated with one student in between them is

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