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Step-by-Step Solution
Step 1: Identify the letters and their frequencies
The word “MATHEMATICS” contains 11 letters:
M, A, T, H, E, M, A, T, I, C, S.
− M appears 2 times
− A appears 2 times
− T appears 2 times
− H, E, I, C, and S each appear 1 time
Step 2: Total number of distinct arrangements
First, calculate the total number of ways to arrange all 11 letters (considering the repeated letters). This is given by dividing 11! by the product of factorials of the frequencies. Hence:
$ \displaystyle \frac{11!}{2!\,2!\,2!} $
Here, each 2! in the denominator accounts for the fact that rearranging two identical letters among themselves does not produce a new arrangement.
Step 3: Number of arrangements with C and S together
To find how many arrangements have C and S together, treat “C” and “S” as a single block. Within that block, they can be arranged in 2! ways (CS or SC).
• Now we effectively have 10 ‘units’: the block (CS) plus the other 9 distinct letters (M, A, T, H, E, M, A, T, I).
• Among the 9 letters (excluding the CS block), M, A, and T each appear twice.
The number of ways to arrange these 10 units is:
$ \displaystyle \frac{10!}{2!\,2!\,2!} $
Then multiply by 2! to account for the internal arrangement of “C” and “S” within that block:
$ \displaystyle \frac{10!}{2!\,2!\,2!} \times 2! = \frac{10!}{2!\,2!} \,.$
Step 4: Number of arrangements where C and S do not come together
We subtract the count from Step 3 (C and S together) from the total count in Step 2 (all possible arrangements):
$ \displaystyle
\left(\frac{11!}{2!\,2!\,2!}\right)
\;-\;
\left(\frac{10!}{2!\,2!}\right) \,.
$
Step 5: Simplify to find the expression in terms of $(6!)$
Factor out $ \displaystyle \frac{10!}{2!\,2!} $:
$ \displaystyle
\frac{10!}{2!\,2!}
\left[
\frac{11}{2!} - 1
\right]
=
\frac{10!}{2!\,2!}
\left[
\frac{11}{2} - 1
\right]
=
\frac{10!}{2!\,2!}
\times
\frac{9}{2}.
$
Further simplification leads to a final count of
$ 6! \times 5670 $.
Step 6: Identify the value of $k$
According to the problem, the total number of such arrangements is written as
$ (6!) \, k $. Matching this with $ 6! \times 5670 $ shows:
$ \displaystyle k = 5670 \,.$
Final Answer
5670
