The word "REPOSITORY" consists of the following letters:

• R (appears 2 times)
• E (appears 1 time)
• P (appears 1 time)
• O (appears 2 times)
• S (appears 1 time)
• I (appears 1 time)
• T (appears 1 time)
• Y (appears 1 time)

We need to arrange the letters such that all identical letters are adjacent to each other but no two groups of identical letters are adjacent to each other. The solution involves several steps:

Step 1: Treat each group of identical letters as a single unit

First, group the identical letters together:

• RR (group of two R's)
• OO (group of two O's)
• E, P, S, I, T, Y (all are unique)

This gives us the following units to arrange:

• RR, OO, E, P, S, I, T, Y

In total, we have 8 units.

Step 2: Arrange the distinct groups (units)

There are 8 distinct units (RR, OO, E, P, S, I, T, Y), which can be arranged in $8!$ ways.

$8! = 40320$

Step 3: Ensure no two groups of identical letters are adjacent

Now, we need to ensure that the two groups of identical letters (RR and OO) are not adjacent. To do this, we first arrange the 6 units that don't involve the identical letters (E, P, S, I, T, Y). These 6 units can be arranged in $6!$ ways.

$6! = 720$

After arranging these 6 units, there are 7 possible spaces where we can place the groups RR and OO (before the first unit, between units, or after the last unit). To place the two groups in these 7 spaces such that they are not adjacent, we can select 2 spaces out of the 7, which can be done in $\binom{7}{2}$ ways.

$\binom{7}{2} = 21$

Step 4: Final computation

Since we can place the two groups (RR and OO) in any of the 21 ways, the total number of valid arrangements is:

$6! \times \binom{7}{2} = 720 \times 21 = 15120$

Thus, the total number of ways to arrange the letters in "REPOSITORY" such that all identical letters are adjacent to each other but no two groups of identical letters are adjacent to each other is $\boxed{15120}$.

Would you like more details or have any questions? Here are some related questions:

1. What happens if all the letters in "REPOSITORY" were distinct?
2. How would the solution change if there were three identical letters instead of two?
3. How do binomial coefficients relate to this type of arrangement problem?
4. What if we allowed the groups of identical letters to be adjacent to each other?
5. How would the solution change if there were more than two groups of identical letters?

Tip: When solving problems involving letter arrangements with constraints, grouping identical items and considering spaces to avoid adjacency is a key strategy.