On her vacations,Veena visits four cities (A, | vedclass

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(B) The total number of ways to visit $4$ cities in a random order is $4! = 4 	imes 3 	imes 2 	imes 1 = 24$.Let $G$ be the event that she visits $A

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$1$/$12$

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(B) The total number of ways to visit $4$ cities in a random order is $4! = 4 	imes 3 	imes 2 	imes 1 = 24$.Let $G$ be the event that she visits $A$ first and $B$ last.If $A$ is fixed at the first position and $B$ is fixed at the last position,the remaining $2$ cities ($C$ and $D$) can be arranged in the middle $2$ positions in $2! = 2$ ways.These arrangements are $(A, C, D, B)$ and $(A, D, C, B)$.Thus,the number of favorable outcomes $n(G) = 2$.The probability $P(G)$ is given by:$P(G) = \frac{n(G)}{n(S)} = \frac{2}{24} = \frac{1}{12}$.

On her vacations,Veena visits four cities ($A, B, C$,and $D$) in a random order. What is the probability that she visits $A$ first and $B$ last?

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On her vacations,Veena visits four cities ($A, B, C$,and $D$) in a random order. What is the probability that she visits $A$ before $B$ and $B$ before $C$?

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