If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which C and S do not come together, is $$(6!)k$$ then $$k$$ is equal to

Given: The word MATHEMATICS has 11 letters: M, A, T, H, E, M, A, T, I, C, S.

The repeated letters are: M appears 2 times, A appears 2 times, T appears 2 times. The remaining letters H, E, I, C, S each appear once.

Total arrangements:

$$\text{Total} = \frac{11!}{2! \times 2! \times 2!} = \frac{39916800}{8} = 4989600$$

Arrangements where C and S are together:

Treat C and S as a single unit. This gives 10 items to arrange (with M appearing 2 times, A appearing 2 times, T appearing 2 times). The unit CS can be arranged internally in $$2! = 2$$ ways (CS or SC).

$$\text{Together} = \frac{10!}{2! \times 2! \times 2!} \times 2! = \frac{3628800}{8} \times 2 = 453600 \times 2 = 907200$$

Arrangements where C and S do NOT come together:

$$\text{Not together} = 4989600 - 907200 = 4082400$$

We are given that this equals $$(6!)k$$:

$$k = \frac{4082400}{6!} = \frac{4082400}{720} = 5670$$

The correct answer is Option B: 5670.