Q102 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
In how many ways can the word "CHRISTMAS" be arranged so that the letters '$\mathrm{C}$' and '$\mathrm{M}$' are never adjacent?
- A$9!\times \dfrac{7}{2}$
- B$8!\times \dfrac{9}{2}$
- Correct answerC$8!\times \dfrac{7}{2}$
- D$7!\times \dfrac{9}{2}$
Solution
The word CHRISTMAS contains 9 letters: C, H, R, I, S, T, M, A, S. The letter S repeats twice. The total number of arrangements is $\dfrac{9!}{2!}$.
To find the number of arrangements where C and M are never adjacent, we subtract the arrangements where C and M are adjacent from the total number of arrangements.
Treating (CM) as a single unit, we have 8 units: (CM), H, R, I, S, T, A, S. The number of arrangements of these 8 units is $\dfrac{8!}{2!}$. Since C and M can be arranged as CM or MC, we multiply by 2!.
Number of arrangements where C and M are adjacent = $\dfrac{8!}{2!} \times 2! = 8!$.
Number of arrangements where C and M are never adjacent = $\dfrac{9!}{2!} - 8! = \dfrac{9 \times 8!}{2} - 8! = 8! \times (\dfrac{9}{2} - 1) = 8! \times \dfrac{7}{2}$.
Answer: $8!\times \dfrac{7}{2}$
To find the number of arrangements where C and M are never adjacent, we subtract the arrangements where C and M are adjacent from the total number of arrangements.
Treating (CM) as a single unit, we have 8 units: (CM), H, R, I, S, T, A, S. The number of arrangements of these 8 units is $\dfrac{8!}{2!}$. Since C and M can be arranged as CM or MC, we multiply by 2!.
Number of arrangements where C and M are adjacent = $\dfrac{8!}{2!} \times 2! = 8!$.
Number of arrangements where C and M are never adjacent = $\dfrac{9!}{2!} - 8! = \dfrac{9 \times 8!}{2} - 8! = 8! \times (\dfrac{9}{2} - 1) = 8! \times \dfrac{7}{2}$.
Answer: $8!\times \dfrac{7}{2}$
Hence, the correct answer is option: 'C'.