Count total arrangements of the letters of $\text{MATHEMATICS}$.

The word $\text{MATHEMATICS}$ has 11 letters: $M, A, T, H, E, M, A, T, I, C, S$

Repeated letters are:

• $M$ occurs 2 times
• $A$ occurs 2 times
• $T$ occurs 2 times

Hence total number of distinct arrangements is $\frac{11!}{2!\,2!\,2!} = \frac{11!}{8}.$

Count arrangements in which $C$ and $S$ come together.

Treat $C$ and $S$ as one block. Since they can appear as $CS$ or $SC$, there are $2$ internal arrangements.

Now we arrange:

• the block $(CS)$ or $(SC)$
• $M, M$
• $A, A$
• $T, T$
• $H, E, I$

Total objects $= 10$.

With repetitions of $M, A, T$, the number of such arrangements is $2\cdot \frac{10!}{2!\,2!\,2!} = 2\cdot \frac{10!}{8} = \frac{10!}{4}.$

Count arrangements in which $C$ and $S$ do not come together.

Required number $= \frac{11!}{8} - \frac{10!}{4}.$

Factor out $10!$:

Express in the form $(6!)k$.

Since $10! = 10\cdot 9\cdot 8\cdot 7\cdot 6!,$ we get $\frac{9\cdot 10!}{8} = \frac{9\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6!}{8}.$

Cancel $8$: $= 9\cdot 10\cdot 9\cdot 7\cdot 6!.$

Therefore $k = 9\cdot 10\cdot 9\cdot 7 = 5670.$

Check options

• A: $5670$ ✅
• B: $1890$
• C: $2835$
• D: $945$

So the correct option is A.