if letters of the word MATHEMATICS are arranged then the probability that C come before E,E before H ,H before I and I before S

Options

A. `(3)/(10)`
B. `(1)/(20)`
C. `(1)/(120)`
D. `(1)/(720)`

Answer

To solve the problem, we need to find the probability that in the arrangement of the letters of the word "MATHEMATICS", the letters C, E, H, I, and S appear in the order C before E, E before H, H before I, and I before S.

### Step-by-step Solution:

1. **Count the Total Letters:**
The word "MATHEMATICS" has 11 letters: M, A, T, H, E, M, A, T, I, C, S.

2. **Identify Repeated Letters:**
The letters M, A, and T are repeated:
- M appears 2 times
- A appears 2 times
- T appears 2 times

3. **Calculate Total Arrangements:**
The total number of arrangements of the letters in "MATHEMATICS" can be calculated using the formula for permutations of multiset:
\[ \text{Total arrangements} = \frac{11!}{2! \times 2! \times 2!} \]

4. **Fix the Order of C, E, H, I, S:**
We need to ensure that C comes before E, E comes before H, H comes before I, and I comes before S. This means that the relative order of these letters is fixed.

5. **Choose Positions for C, E, H, I, S:**
We need to choose 5 positions out of the 11 for the letters C, E, H, I, and S. The number of ways to choose 5 positions from 11 is given by:
\[ \binom{11}{5} \]

6. **Arrange Remaining Letters:**
After placing C, E, H, I, and S, we have 6 positions left for the remaining letters (M, A, T, M, A, T). The arrangements of these letters can be calculated as:
\[ \frac{6!}{2! \times 2! \times 2!} \]

7. **Calculate the Probability:**
The probability that C comes before E, E before H, H before I, and I before S is given by the ratio of the favorable arrangements to the total arrangements:
\[ P = \frac{\text{Number of favorable arrangements}}{\text{Total arrangements}} = \frac{\binom{11}{5} \cdot \frac{6!}{2! \times 2! \times 2!}}{\frac{11!}{2! \times 2! \times 2!}} \]

8. **Simplifying the Expression:**
The \(2! \times 2! \times 2!\) cancels out from the numerator and denominator:
\[ P = \frac{\binom{11}{5} \cdot 6!}{11!} \]
Since \(11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6!\), we can simplify further:
\[ P = \frac{\binom{11}{5}}{11 \times 10 \times 9 \times 8 \times 7} \]

9. **Calculating \( \binom{11}{5} \):**
\[ \binom{11}{5} = \frac{11!}{5! \cdot 6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462 \]

10. **Final Calculation of Probability:**
\[ P = \frac{462}{11 \times 10 \times 9 \times 8 \times 7} = \frac{462}{55440} = \frac{1}{120} \]

### Final Answer:
The probability that C comes before E, E before H, H before I, and I before S is \( \frac{1}{120} \).