Key Concepts and Formulas

• Permutations with Repetitions: The number of distinct permutations of $n$ objects, where $n_1$ are of one type, $n_2$ are of another type, ..., $n_k$ are of the $k$-th type, is given by: $\frac{n!}{n_1! n_2! \dots n_k!}$
• Complementary Counting: The number of ways to satisfy a condition is equal to the total number of possibilities minus the number of ways the condition is not satisfied. $\text{Number of ways (condition satisfied)} = \text{Total number of ways} - \text{Number of ways (condition not satisfied)}$

Step-by-Step Solution

1. Analyze the word and identify repetitions. The given word is MATHEMATICS. We identify the frequency of each letter:

• M: 2
• A: 2
• T: 2
• H: 1
• E: 1
• I: 1
• C: 1
• S: 1 The total number of letters is $n = 11$.

2. Calculate the total number of arrangements without any restrictions. Using the permutations with repetitions formula, we have: $\text{Total arrangements} = \frac{11!}{2!2!2!}$ This is because there are 11 letters in total, with M, A, and T each appearing twice. $\text{Total arrangements} = \frac{39916800}{8} = 4989600$

3. Calculate the number of arrangements where C and S are together. We treat CS as a single unit. This reduces the number of items to arrange to 10: M, A, T, H, E, I, (CS), M, A, T. Now we have 10 items with repetitions: M (2 times), A (2 times), T (2 times). The number of arrangements is: $\frac{10!}{2!2!2!}$ Since C and S can be arranged as CS or SC, we multiply by $2!$: $\text{Arrangements with C and S together} = \frac{10!}{2!2!2!} \times 2! = \frac{10!}{2!2!} = \frac{3628800}{4} = 907200$

4. Apply the principle of complementary counting. We want to find the number of arrangements where C and S are not together. Using complementary counting: $\text{Arrangements with C and S not together} = \text{Total arrangements} - \text{Arrangements with C and S together}$ $\text{Arrangements with C and S not together} = \frac{11!}{2!2!2!} - \frac{10!}{2!2!}$ $= \frac{11 \cdot 10!}{8} - \frac{10!}{4} = \frac{11 \cdot 10!}{8} - \frac{2 \cdot 10!}{8} = \frac{9 \cdot 10!}{8}$

5. Express the result in the form $(6!)k$ and find the value of $k$. We simplify the expression: $\frac{9 \cdot 10!}{8} = \frac{9 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6!}{8} = 9 \cdot 10 \cdot 9 \cdot 7 \cdot 6! = 5670 \cdot 6!$ We are given that the number of words is $(6!)k$. Comparing $5670 \cdot 6!$ with $(6!)k$, we find: $k = 5670$

Tips and Common Mistakes to Avoid

• Double-check Repetitions: Always verify the number of repetitions of each letter in the word. A small error here propagates through the entire calculation.
• Don't forget Internal Arrangements: When grouping letters together, remember to account for the different ways the letters within the group can be arranged.
• Complementary Counting is Key: Recognizing when to use complementary counting can significantly simplify the problem. Trying to directly count the arrangements where C and S are not together is much more complex.

Summary

We calculated the total number of arrangements of the letters in MATHEMATICS, then subtracted the number of arrangements where C and S are together, using the principle of complementary counting. We then expressed the result in the form $(6!)k$ to find the value of $k$. The value of $k$ is 5670.

The final answer is \boxed{5670}, which corresponds to option (A).