Letters are arranged in all possible ways of the words `DISCRETE MATHMATICS'. One word is selected at random then the probability that all T are at maximum distance is a/b ( a and b has no common factor) then a is_____.

Answer

To solve the problem, we need to find the probability that all T's in the phrase "DISCRETE MATHMATICS" are at maximum distance from each other when the letters are arranged in all possible ways.

### Step-by-Step Solution:

1. **Count the Total Letters:**
The phrase "DISCRETE MATHMATICS" has 15 letters: D, I, S, C, R, E, T, E, M, A, T, H, M, A, T, I, C, S.
- Count of letters:
- D: 1
- I: 2
- S: 2
- C: 2
- R: 1
- E: 2
- M: 2
- A: 2
- T: 3

2. **Calculate Total Arrangements:**
The total arrangements of the letters can be calculated using the formula for permutations of multiset:
\[ \text{Total arrangements} = \frac{15!}{(2!)^6 \cdot 3!} \]
Here, \(2!\) is for each letter that appears twice (I, S, C, E, M, A) and \(3!\) is for T which appears thrice.

3. **Arranging T's at Maximum Distance:**
To ensure that all T's are at maximum distance, we can place the T's in such a way that there are 7 letters between them.
- We can visualize it as: T _ _ _ _ _ _ _ T _ _ _ _ _ _ T
- This arrangement requires that we have 7 letters between each T.

4. **Calculate Arrangements with T's at Maximum Distance:**
After placing the T's, we have 12 remaining letters (D, I, S, C, R, E, M, A, I, C, S) to arrange in the remaining slots.
- The arrangement of these 12 letters is:
\[ \text{Arrangements with T's at max distance} = \frac{12!}{(2!)^5} \]
(Here \(2!\) is for letters that appear twice: I, S, C, E, M, A).

5. **Calculate the Probability:**
The probability that all T's are at maximum distance is given by the ratio of the favorable arrangements to the total arrangements:
\[ P(\text{T's at max distance}) = \frac{\frac{12!}{(2!)^5}}{\frac{15!}{(2!)^6 \cdot 3!}} \]
Simplifying this gives:
\[ P = \frac{12! \cdot (2!)^6 \cdot 3!}{15! \cdot (2!)^5} \]
\[ = \frac{12! \cdot 2! \cdot 3!}{15!} \]
\[ = \frac{1}{3 \cdot 17 \cdot 4} \]

6. **Final Probability:**
The final probability can be expressed as \( \frac{1}{204} \) which is in the form \( \frac{a}{b} \) where \( a = 1 \) and \( b = 204 \).

### Conclusion:
Thus, the value of \( a \) is **1**.