Discussion

Explanation:

Let us first arrange the other 10 letters.

∴ Number of ways of arranging these letters = $\frac{10!}{2!\times 2!\times 2!}$.

Now, we have 11 spaces created to put the three I’s.

We can select any three of these in ¹¹C₃ = 165 ways.

∴ Total number of words that can be formed such that no two ‘I’s are adjacent to each other = 165 × $\frac{10!}{2!\times \times 2!\times 2!}$.

Hence, option (a).

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