The number of ways in which the letters AAABBCD can be arranged so that two Bs are together but no two As are together is:

Steps to Solve

1. Treat the two B's as a single unit Combine the two B's into one unit, calling it "BB". The letters to arrange are now: A, A, A, C, D, BB. This gives us 6 units total.

2. Calculate arrangements without restrictions Calculate the arrangements of the 6 units (A, A, A, C, D, BB). The formula for arrangements of indistinguishable objects is:

$$ \text{Arrangements} = \frac{n!}{n_1! \times n_2! \times \ldots} $$

Here, we have 3 A's, 1 C, 1 D, and 1 BB:

$$ \text{Arrangements} = \frac{6!}{3! \times 1! \times 1! \times 1!} $$

3. Calculate the total arrangements Calculate the factorial values:

$$ 6! = 720, \quad 3! = 6 $$

Now substitute these into the formula:

$$ \text{Arrangements} = \frac{720}{6} = 120 $$

4. Arrange the A's with restrictions Now, we need to arrange the A's so that no two A's are adjacent.

5. Place non-A letters First, arrange the letters without A's: (C, D, BB) gives us:

$$ \text{Arrangements without A's} = 3! = 6 $$

6. Identify slots for A's When the letters (C, D, BB) are arranged, they create gaps to place the A's. There are 4 gaps: before C, between C and D, between D and BB, and after BB.

7. Place A's in slots We need to place 3 A's in these 4 slots where no two A's are together. The number of ways to choose 3 slots from the available 4 slots is given by:

$$ \binom{4}{3} = 4 $$

8. Combine results Now, multiply the arrangements of non-A letters by the combinations of A's placement:

$$ \text{Total arrangements} = 6 \times 4 = 24 $$