Q87 / 180 · COMEDK 2022 · Mathematics
Five persons $A, B, C, D$ and $E$ are in queue of a shop. The Probability that $A$ and $B$ are always together is
- Correct answerA$\dfrac{2}{5}$
- B$\dfrac{1}{4}$
- C$\dfrac{3}{5}$
- D$\dfrac{2}{3}$
Solution
The total number of ways to arrange 5 persons $A, B, C, D, E$ in a queue is $5! = 120$.
To find the number of arrangements where $A$ and $B$ are always together, treat $(AB)$ as a single unit. Now, we have 4 units: $(AB), C, D, E$.
These 4 units can be arranged in $4! = 24$ ways. Within the unit $(AB)$, $A$ and $B$ can be arranged in $2! = 2$ ways ($AB$ or $BA$).
Therefore, the total number of favorable arrangements is $24 \times 2 = 48$.
The probability is the ratio of favorable arrangements to total arrangements:
$P = \dfrac{48}{120} = \dfrac{2}{5}$.
Answer: $\dfrac{2}{5}$
To find the number of arrangements where $A$ and $B$ are always together, treat $(AB)$ as a single unit. Now, we have 4 units: $(AB), C, D, E$.
These 4 units can be arranged in $4! = 24$ ways. Within the unit $(AB)$, $A$ and $B$ can be arranged in $2! = 2$ ways ($AB$ or $BA$).
Therefore, the total number of favorable arrangements is $24 \times 2 = 48$.
The probability is the ratio of favorable arrangements to total arrangements:
$P = \dfrac{48}{120} = \dfrac{2}{5}$.
Answer: $\dfrac{2}{5}$
Hence, the correct answer is option: 'A'.