Question 1210171
We are asked to count the number of ways to place **three A's, three B's, and three C's** in a **3×3 grid** such that:

- Every square contains exactly one letter.
- Each **row** and **column** ends up with 3 letters total (since the grid is filled).
- **Each diagonal** (main and anti-diagonal) must contain exactly **one A, one B, and one C**.

---

### Step 1: Total number of arrangements without the diagonal condition

We need to place 3 A’s, 3 B’s, and 3 C’s into 9 positions. This is a classic **multinomial coefficient**:

\[
\frac{9!}{3! \cdot 3! \cdot 3!} = \frac{362880}{6 \cdot 6 \cdot 6} = \frac{362880}{216} = 1680
\]

So there are **1680** total ways to fill the grid with 3 A’s, 3 B’s, and 3 C’s.

---

### Step 2: Apply the diagonal condition

We must now count **only those** of the 1680 arrangements in which:

- The **main diagonal** (top-left to bottom-right) contains one A, one B, and one C.
- The **anti-diagonal** (top-right to bottom-left) also contains one A, one B, and one C.

Let’s define the positions in the grid like this:

```
(1,1) (1,2) (1,3)
(2,1) (2,2) (2,3)
(3,1) (3,2) (3,3)
```

The **main diagonal** is: (1,1), (2,2), (3,3)  
The **anti-diagonal** is: (1,3), (2,2), (3,1)

Note that the **center square (2,2)** is on **both diagonals**.

---

So the union of the two diagonals has 5 distinct squares:

- Main: (1,1), (2,2), (3,3)
- Anti: (1,3), (2,2), (3,1)

=> Union: **(1,1), (2,2), (3,3), (1,3), (3,1)**

This gives us **5 positions**, with **(2,2)** counted once (in both diagonals).

---

We want these two diagonals (i.e., the five cells) to include:

- Each of A, B, and C **exactly once** in each diagonal.

Let’s approach it step by step.

---

### Step 3: Count the valid diagonal assignments

We'll now count the number of ways to assign letters to the **5 diagonal positions**, so that:

- (1,1), (2,2), (3,3) have one A, one B, one C (i.e., main diagonal is a permutation of A, B, C)
- (1,3), (2,2), (3,1) also have one A, one B, one C (anti-diagonal is a permutation of A, B, C)

Key insight: The **center** cell (2,2) is in **both** diagonals, so its letter is shared.

Let’s choose the letter for the **center cell (2,2)** first.

We have **3 choices**: A, B, or C.

For each such choice:

- The main diagonal now needs to use the other two letters (excluding the center), and arrange them in the remaining two diagonal spots: (1,1) and (3,3).
- Similarly, the anti-diagonal uses the same idea for (1,3) and (3,1).

So for each choice of letter at (2,2), the number of ways to assign letters to the rest of the diagonals is:

\[
(\text{ways to fill the other 2 positions in main diagonal}) \times (\text{ways to fill the other 2 in anti-diagonal}) = 2! \times 2! = 2 \times 2 = 4
\]

So total number of valid ways to fill the **5 diagonal squares** is:

\[
3 \text{ (choices for center)} \times 4 = 12
\]

---

### Step 4: Fill the remaining 4 squares

Now we’ve filled 5 of the 9 squares, each with one letter (from A, B, or C). We used 5 letters total, one per square.

So we have 4 squares left, and **we need to use up the remaining letters** so that each letter (A, B, C) appears exactly 3 times.

Let’s do the math:

We used:
- 1 A, 1 B, 1 C in each diagonal (so 2 of each total),
- But the center letter was **shared**, so total count so far is:

  - The center letter appears once.
  - The other two letters appear twice each (once in each diagonal).

Let’s break it down depending on the center letter. For example:

- Suppose the center is A.
- Then main diagonal: A (center), B, C → uses 1 B, 1 C
- Anti-diagonal: A (same center), B, C → uses another 1 B, 1 C

So we’ve placed:
- A: 1 time
- B: 2 times
- C: 2 times

So we have:
- 2 more A’s to place
- 1 more B
- 1 more C

Total of 4 letters to place in the 4 remaining squares.

This pattern is the same no matter what the center letter is — we always end up needing to place **2 of one letter** (the center one) and **1 of each of the other two**.

So for each of the 12 diagonal configurations, the remaining 4 letters to place are fixed, and we need to count the number of ways to assign them to the 4 leftover squares.

We are placing 4 letters: one letter appears twice, the others once each.

So the number of arrangements is:

\[
\frac{4!}{2! \cdot 1! \cdot 1!} = 12
\]

---

### Step 5: Multiply it all together

- 12 valid ways to assign letters to the 5 diagonal squares  
- For each of those, 12 ways to fill the other 4 squares

So total number of valid arrangements is:

\[
12 \times 12 = \boxed{144}
\]

---

### ✅ Final Answer:

\[
\boxed{144}
\]