1. Count the letters in $\text{MISSISSIPPI}$

The word $\text{MISSISSIPPI}$ has:

• $M = 1$
• $I = 4$
• $S = 4$
• $P = 2$

We need the number of distinct arrangements in which no two S are adjacent.

2. First arrange all letters except S

Remove the $4$ S's. Then we arrange the letters: $M, I, I, I, I, P, P$

Total number of distinct arrangements of these $7$ letters is $\frac{7!}{4!2!} = 105$

3. Create gaps for placing S's

For any arrangement of these $7$ letters, there are gaps where S can be placed so that no two S are adjacent:

$\_\; a \;\_\; b \;\_\; c \;\_\; d \;\_\; e \;\_\; f \;\_\; g \;\_$

Since there are $7$ arranged letters, the number of gaps is $7+1=8$

To ensure no two S's are adjacent, we place the $4$ identical S's in 4 distinct gaps chosen from these $8$ gaps.

Number of ways: $\binom{8}{4}$

4. Total number of valid arrangements

Thus total number of words is $\frac{7!}{4!2!}\binom{8}{4}$

Now, $\frac{7!}{4!2!} = \frac{7\cdot 6\cdot 5!}{(4\cdot 3\cdot 2\cdot 1)(2)(5!)} = \frac{7\cdot 6\cdot 5}{24\cdot 2} = 105$

Also, $105 = 7\cdot 15 = 7\binom{6}{4}$ because $\binom{6}{4}=15$

So the expression becomes $7\binom{6}{4}\binom{8}{4}$

This matches Option D.

5. Check options

• A: $8\binom{6}{4}\binom{7}{4}$ — not equal to the required count.
• B: $6\cdot 7\binom{8}{4}$ — missing the correct combinatorial factor.
• C: $6\cdot 8\binom{7}{4}$ — incorrect.
• D: $7\binom{6}{4}\binom{8}{4}$ — correct.

Therefore, the correct option is $\boxed{D}$