A,A,B,B,C,C,D,E,F are arranged in a row so that no two alike alphabets are together. Find number of such arrangment

Answer

To solve the problem of arranging the letters A, A, B, B, C, C, D, E, F such that no two alike letters are together, we can follow these steps:

### Step 1: Calculate the total arrangements without restrictions
First, we need to find the total arrangements of the letters without any restrictions. The total number of letters is 9, with A, B, and C each appearing twice.

The formula for the arrangements of n objects where some objects are identical is given by:

\[ \text{Total arrangements} = \frac{n!}{n_1! \cdot n_2! \cdot n_3!} \]

Where:
- \( n \) is the total number of letters,
- \( n_1, n_2, n_3 \) are the frequencies of the identical letters.

In our case:

\[ \text{Total arrangements} = \frac{9!}{2! \cdot 2! \cdot 2!} \]

Calculating this gives:

\[ 9! = 362880 \quad \text{and} \quad 2! = 2 \]

So,

\[ \text{Total arrangements} = \frac{362880}{2 \cdot 2 \cdot 2} = \frac{362880}{8} = 45360 \]

### Step 2: Calculate arrangements where at least two identical letters are together
Next, we will use the principle of inclusion-exclusion to find the number of arrangements where at least two identical letters are together.

Let:
- \( A \) be the set of arrangements where the A's are together.
- \( B \) be the set of arrangements where the B's are together.
- \( C \) be the set of arrangements where the C's are together.

We want to find \( |A \cup B \cup C| \).

Using the principle of inclusion-exclusion:

\[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| \]

### Step 3: Calculate \( |A|, |B|, |C| \)
If we consider A's as one unit (AA), we have the following letters to arrange: AA, B, B, C, C, D, E, F (8 units).

The number of arrangements is:

\[ |A| = \frac{8!}{2! \cdot 2!} = \frac{40320}{4} = 10080 \]

Similarly, \( |B| \) and \( |C| \) will also be:

\[ |B| = |C| = 10080 \]

### Step 4: Calculate \( |A \cap B|, |A \cap C|, |B \cap C| \)
If both A's and B's are together, we treat them as units (AA and BB). The letters to arrange are: AA, BB, C, C, D, E, F (7 units).

The number of arrangements is:

\[ |A \cap B| = |A \cap C| = |B \cap C| = \frac{7!}{2!} = \frac{5040}{2} = 2520 \]

### Step 5: Calculate \( |A \cap B \cap C| \)
If A's, B's, and C's are all together, we treat them as units (AA, BB, CC). The letters to arrange are: AA, BB, CC, D, E, F (6 units).

The number of arrangements is:

\[ |A \cap B \cap C| = 6! = 720 \]

### Step 6: Substitute into the inclusion-exclusion formula
Now we can substitute these values into the inclusion-exclusion formula:

\[ |A \cup B \cup C| = 10080 + 10080 + 10080 - 2520 - 2520 - 2520 + 720 \]

Calculating this gives:

\[ |A \cup B \cup C| = 30240 - 7560 + 720 = 23040 \]

### Step 7: Calculate arrangements where no two identical letters are together
Finally, the number of arrangements where no two identical letters are together is given by:

\[ \text{Arrangements with no identical letters together} = \text{Total arrangements} - |A \cup B \cup C| \]

Substituting the values we found:

\[ \text{Arrangements with no identical letters together} = 45360 - 23040 = 22320 \]

Thus, the final answer is:

\[ \boxed{22320} \]