Question

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 , is equal to

• A. 5670
• B. 1890
• C. 595
• D. 657

Answer 1

The Correct Option is A

1) Letter multiset

MATHEMATICS (11 letters):
\(M^2,\, A^2,\, T^2,\, H,\, E,\, I,\, C,\, S\).

2) Total arrangements (no restriction)

\[ N_{\text{all}}=\frac{11!}{2!\,2!\,2!}=\frac{11!}{8}. \]

3) Arrangements with C and S together

Treat the block \(\boxed{CS}\) (or \(\boxed{SC}\)) as one item. Then we have 10 items: the block \(+\) \(M^2,A^2,T^2,H,E,I\).
External permutations: \[ \frac{10!}{2!\,2!\,2!}=\frac{10!}{8},\quad \text{and internal choices for the block }(CS/SC)=2!. \] Hence \[ N_{\text{together}}=2\cdot \frac{10!}{8}=\frac{10!}{4}. \]

4) Required count (not together)

\[ N = N_{\text{all}}-N_{\text{together}} = \frac{11!}{8}-\frac{10!}{4} = \frac{10!}{8}\left(11-2\right) = \frac{9}{8}\,10!. \]

5) Express as \((6!)\,k\)

Write \(10!=6!\cdot 7\cdot 8\cdot 9\cdot 10\). Then \[ N=\frac{9}{8}\,10! =\frac{9}{8}\,(6!\cdot 7\cdot 8\cdot 9\cdot 10) =6!\,\big( \tfrac{9}{8}\cdot 7\cdot 8\cdot 9\cdot 10 \big) =6!\,(7\cdot 9\cdot 9\cdot 10) =6!\,\underline{5670}. \] Therefore \(k=5670\).

Final: Number of required words \(= (6!)\,\mathbf{5670}\). Hence, \(k=\boxed{5670}\).

Answer 2

The total number of words can be calculated by subtracting the number of words when C and S are together from the total words.
\[ \text{M2A2T2HEICS} = \text{total words} - \text{when C and S are together} \] 

Now, we calculate the total number of words:
\[ \frac{11!}{2!2!2!} - \frac{10!}{2!2!2!} \times 2 \] 

This simplifies to:
\[ \frac{10!}{2!2!} \times 9 = \frac{9 \times 10 \times 9 \times 8 \times 7}{8} \] 

Finally, the result is:
5670