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

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(B) The total number of ways to visit $4$ cities is $4! = 4 	imes 3 	imes 2 	imes 1 = 24$. Let the event $I$ be that she visits $A$ just before $B$

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(B) The total number of ways to visit $4$ cities is $4! = 4 	imes 3 	imes 2 	imes 1 = 24$. Let the event $I$ be that she visits $A$ just before $B$. To find the number of favorable outcomes,we can treat $(AB)$ as a single unit. Now,we have $3$ units: $(AB), C, D$. The number of ways to arrange these $3$ units is $3! = 3 	imes 2 	imes 1 = 6$. The favorable outcomes are: $ABCD, ABDC, CABD, CDAB, DABC, DCAB$. Thus,$n(I) = 6$. The probability $P(I) = \frac{n(I)}{n(S)} = \frac{6}{24} = \frac{1}{4}$.

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

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