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

Concept: Classical probability on permutations. When all orderings are equally likely, count favorable arrangements and divide by total arrangements.

Total arrangements of four cities: $4! = 24$.

1. A before B: In any arrangement of A and B among the four positions, exactly half have A before B by symmetry. Probability $= \frac{1}{2}$.
2. A before B and B before C: Among the $3! = 6$ orderings of A, B, C (ignoring D's position), only one is A–B–C. Probability $= \frac{1}{6}$.
3. A first and B last: Fix A in position 1, B in position 4; arrange C, D in the middle two positions: $2! = 2$ ways. Probability $= \frac{2}{24} = \frac{1}{12}$.
4. A either first or second: A first: $3! = 6$ arrangements. A second: $3! = 6$ arrangements. Total $= 12$. Probability $= \frac{12}{24} = \frac{1}{2}$.
5. A just before B (adjacent, A–B): Treat A–B as a single block. Arrange 3 units (A–B, C, D): $3! = 6$ ways. Probability $= \frac{6}{24} = \frac{1}{4}$.
   ✓Final answer

   1. $\boxed{\frac{1}{2}}$,
   2. $\boxed{\frac{1}{6}}$,
   3. $\boxed{\frac{1}{12}}$,
   4. $\boxed{\frac{1}{2}}$,
   5. $\boxed{\frac{1}{4}}$.

The heart of this problem is symmetry in permutations. When Veena visits four cities in a random order, all $4! = 24$ orderings are equally likely. For any subset of cities, every relative ordering of that subset appears equally often across all permutations. This symmetry principle lets us count favorable outcomes without listing every arrangement.

The total number of ways to visit four cities is $4! = 24$.

(i) Probability that A comes before B

Among all orderings of four cities, consider just the relative positions of A and B. By symmetry, in exactly half of all arrangements A appears before B, and in the other half B appears before A. The presence of C and D doesn't break this symmetry—they simply fill the remaining slots.

The probability is $\frac{12}{24} = \frac{1}{2}$.

(ii) Probability that A before B and B before C

Now we need the specific ordering A, then B, then C (with D anywhere). Among the three cities A, B, C, there are $3! = 6$ possible relative orderings:

• ABC, ACB, BAC, BCA, CAB, CBA

Only one of these six orderings satisfies "A before B before C": namely ABC.

Since all relative orderings of any three cities are equally likely, the probability is $\frac{1}{6}$.

Alternatively, count directly: D can occupy any of the 4 positions. Once D's position is fixed, the remaining 3 positions must be filled by A, B, C in that specific order. There are $4$ ways to place D and $1$ way to arrange A, B, C in the required order, giving $4$ favorable outcomes out of $24$ total: $\frac{4}{24} = \frac{1}{6}$.

The probability is $\frac{1}{6}$.

(iii) Probability that A is first and B is last

We need A in position 1 and B in position 4. The middle two positions (2 and 3) can be filled by C and D in any order.

1. Fix A in position 1: done.
2. Fix B in position 4: done.
3. Arrange C and D in positions 2 and 3: $2! = 2$ ways.

Favorable outcomes: $2$.

The probability is $\frac{2}{24} = \frac{1}{12}$.

(iv) Probability that A is either first or second

We split into two disjoint cases:

Case 1: A is first.

The remaining three cities B, C, D can be arranged in positions 2, 3, 4 in $3! = 6$ ways.

Case 2: A is second.

One of B, C, D occupies position 1 ($3$ choices), and the remaining two cities fill positions 3 and 4 ($2! = 2$ ways each). Total: $3 \times 2 = 6$ ways.

Favorable outcomes: $6 + 6 = 12$.

The probability is $\frac{12}{24} = \frac{1}{2}$.

(v) Probability that A is immediately before B

"A just before B" means A and B occupy consecutive positions with A first. The possible consecutive pairs of positions are:

• (1, 2), (2, 3), (3, 4)

That's 3 pairs of consecutive positions.

For each pair:

1. Place A in the first position of the pair and B in the second.
2. Arrange the remaining two cities (C and D) in the remaining two positions: $2! = 2$ ways.

Favorable outcomes: $3 \times 2 = 6$.

The probability is $\frac{6}{24} = \frac{1}{4}$.