In how many ways can you arrange the letters of the word "PROBABILITY" such that no two I's are adjacent?
To solve this problem, we first need to determine the total number of arrangements for the given word without any restrictions.

The word "PROBABILITY" has 11 letters with the following frequency:
- P: 1
- R: 1
- O: 1
- B: 2
- A: 1
- I: 3
- L: 1
- T: 1
- Y: 1

We can use the multinomial coefficient formula to find the total number of arrangements:

Total arrangements = 11! / (1! * 1! * 1! * 2! * 1! * 3! * 1! * 1! * 1!)
                   = 39916800 / (2 * 6)
                   = 3326400

Now, let's calculate the number of arrangements where at least two I's are adjacent. We can treat the pair of I's as a single unit (II) and arrange the remaining letters:

The new word "PROBABIL(II)TY" has 10 units with the following frequency:
- P: 1
- R: 1
- O: 1
- B: 2
- A: 1
- II: 1
- L: 1
- T: 1
- Y: 1

Total arrangements with at least two I's adjacent = 10! / (1! * 1! * 1! * 2! * 1! * 1! * 1! * 1! * 1!)
                                                 = 3628800 / 2
                                                 = 1814400

Now, we can subtract this number from the total arrangements to find the number of arrangements where no two I's are adjacent:

Arrangements with no two I's adjacent = Total arrangements - Arrangements with at least two I's adjacent
                                     = 3326400 - 1814400
                                     = 1512000

So, there are 1,512,000 ways to arrange the letters of the word "PROBABILITY" such that no two I's are adjacent.