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Step-by-Step Solution
Step 1: Identify the Letters and Their Frequencies
The word “MATHEMATICS” has 11 letters in total. The letters and their counts are:
M – 2 times
A – 2 times
T – 2 times
H – 1 time
E – 1 time
I – 1 time
C – 1 time
S – 1 time
Step 2: Express the Problem of “Not Coming Together”
We need the number of arrangements of these 11 letters in which C and S do not come together. A common strategy is:
Compute the total number of ways to arrange the letters ignoring the restriction.
Compute the number of ways in which C and S are together.
Subtract the number of “C and S together” arrangements from the total number of arrangements.
Alternatively, we can place the remaining letters first and then insert C and S in such a way that they are never adjacent. We will use this second approach.
Step 3: Arrange the Remaining 9 Letters
Remove C and S. The remaining letters are M, A, T, H, E, M, A, T, I, which total 9 letters. Note that M, A, and T each appear twice. The number of ways to arrange these 9 letters is:
$ \frac{9!}{2! \times 2! \times 2!}
= \frac{9!}{2 \times 2 \times 2}
= \frac{362880}{8}
= 45360.
$
Step 4: Count the Available Gaps
When these 9 letters are placed in a row, there are 10 possible “gaps” around them (1 gap before the first letter, 8 gaps between consecutive letters, and 1 gap after the last letter). Symbolically, for 9 letters, the number of gaps is 9 + 1 = 10.
Step 5: Place C and S So That They Are Not Adjacent
We must choose any 2 of these 10 gaps to place C and S, ensuring they are not in adjacent positions. But simply choosing 2 gaps out of 10 inherently ensures they are not adjacent, because each gap is distinct. The number of ways to choose these 2 gaps is:
$ \binom{10}{2} = 45.
$
Moreover, C and S can be arranged among themselves in 2! = 2 ways (C then S, or S then C).
Step 6: Compute the Total Number of Valid Arrangements
Multiplying these factors together gives the number of arrangements where C and S do not come together:
$
\left(\frac{9!}{2!\,2!\,2!}\right) \times \binom{10}{2} \times 2!
= 45360 \times 45 \times 2
= 45360 \times 90
= 4082400.
$
Step 7: Express the Result in Terms of $(6!)k$
We know that $6! = 720.$ The given expression is $(6!)k = 4082400,$ so
$ 4082400 = 720 \times k
\quad\Longrightarrow\quad
k = \frac{4082400}{720}
= 5670.
$
Final Answer
The value of $k$ is 5670.
