To solve the problem of arranging the letters AAAAA, BBB, CCC, D, EE, and F such that the letters B are separated from one another, we can follow these steps:
### Step 1: Count the total letters and arrange them without B
We have the following letters:
- A: 5
- B: 3
- C: 3
- D: 1
- E: 2
- F: 1
First, we will arrange the letters excluding B. The total number of letters excluding B is:
\[ 5 (A) + 3 (C) + 1 (D) + 2 (E) + 1 (F) = 12 \text{ letters} \]
The number of ways to arrange these 12 letters is given by:
\[
\frac{12!}{5! \times 3! \times 2! \times 1!}
\]
### Step 2: Calculate the arrangements of the letters without B
Now, we calculate the value:
\[
12! = 479001600
\]
\[
5! = 120, \quad 3! = 6, \quad 2! = 2, \quad 1! = 1
\]
So,
\[
5! \times 3! \times 2! \times 1! = 120 \times 6 \times 2 \times 1 = 1440
\]
Thus, the number of arrangements of the letters without B is:
\[
\frac{12!}{5! \times 3! \times 2! \times 1!} = \frac{479001600}{1440} = 332640
\]
### Step 3: Identify gaps for placing B
When we arrange the 12 letters, we create gaps where we can place the B's. For 12 letters, there are 13 gaps (one before each letter and one after the last letter).
### Step 4: Choose gaps for B
We need to choose 3 gaps from these 13 to place the B's. The number of ways to choose 3 gaps from 13 is given by:
\[
\binom{13}{3}
\]
Calculating this:
\[
\binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286
\]
### Step 5: Combine the arrangements
Finally, the total number of arrangements of the letters such that the B's are separated is given by multiplying the arrangements of the letters without B by the ways to choose the gaps for B:
\[
\text{Total arrangements} = \frac{12!}{5! \times 3! \times 2! \times 1!} \times \binom{13}{3} = 332640 \times 286
\]
Calculating this gives:
\[
332640 \times 286 = 95193600
\]
### Final Answer
Thus, the total number of ways to arrange the letters AAAAA, BBB, CCC, D, EE, and F such that no two B's are together is:
\[
\boxed{95193600}
\]
