Letters of the word MATHEMATICS are arranged in all the possible ways, and a word is selected randomly then the probability that letter $C$ is exactly between $S$ and $H$ is ______.
Step-by-Step Solution
Key Concept: Treat the three letters S, C, H where C must be between them as a constrained unit with 2 internal arrangements, then count arrangements of remaining units with repetitions.
The word MATHEMATICS has 11 letters: M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1). Total arrangements = $\frac{11!}{2!·2!·2!} = 4989600$. For C to be exactly between S and H, we need the pattern S-C-H or H-C-S in the word. Treat S, C, H as a single unit (2 arrangements: SCH or HCS). Now we arrange 9 units: M(2), A(2), T(2), E(1), I(1), and one (S-C-H) block = $\frac{9!}{2!·2!·2!} × 2 = 181440 × 2 = 362880$. Therefore, probability = $\frac{362880}{4989600} = \frac{1}{13.75} ≈ 0.0727$... Wait, reconsidering: favorable outcomes = $\frac{9!}{2!·2!·2!} × 2 = 90720 × 2 = 181440$. Probability = $\frac{181440}{4989600} = \frac{2}{55} ≈ 0.036363$. Actually the correct calculation gives $\frac{1}{55} = 0.018181$.
Correct Answer: 0.018181