Three a's, three b's and three c's are placed randomly in `3xx3` matrix. The probability that no row or column contain two identical letters can be expressed as `(p)/(q)`, where p and q are coprime then `(p+q)` equals to :

Options

A. 151
B. 161
C. 141
D. 131

Answer

To solve the problem of arranging three A's, three B's, and three C's in a 3x3 matrix such that no row or column contains two identical letters, we will follow these steps:

### Step 1: Calculate Total Arrangements
We start by calculating the total number of arrangements of the letters in the 3x3 matrix. Since we have 3 A's, 3 B's, and 3 C's, the total arrangements can be calculated using the formula for permutations of multiset:

\[ \text{Total arrangements} = \frac{9!}{3! \times 3! \times 3!} \]

### Step 2: Calculate Favorable Outcomes
Next, we need to find the number of favorable outcomes where no row or column contains two identical letters.

To achieve this, we can think of the matrix as having three rows and three columns, where each letter appears exactly once per row and column. This is equivalent to arranging the letters in a way that forms a Latin square.

For a 3x3 matrix, the number of ways to arrange the letters such that no row or column has identical letters can be calculated as follows:

1. Choose a permutation for the first row. There are \(3!\) ways to arrange A, B, C.
2. For the second row, we can choose any arrangement that does not repeat the letters in the same column. There are \(2\) valid arrangements left for the second row.
3. For the third row, only \(1\) arrangement is left that does not repeat the letters in the same column.

Thus, the number of favorable arrangements is:

\[ \text{Favorable outcomes} = 3! \times 2 \times 1 = 6 \]

### Step 3: Calculate the Probability
Now, we can calculate the probability that no row or column contains two identical letters:

\[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total arrangements}} = \frac{6}{\frac{9!}{3! \times 3! \times 3!}} \]

Calculating \(9!\) and \(3!\):

\[ 9! = 362880 \quad \text{and} \quad 3! = 6 \]

So,

\[ 3! \times 3! \times 3! = 6 \times 6 \times 6 = 216 \]

Thus,

\[ \text{Total arrangements} = \frac{362880}{216} = 1680 \]

Now substituting back into the probability formula:

\[ \text{Probability} = \frac{6}{1680} = \frac{1}{280} \]

### Step 4: Expressing the Probability
The probability can be expressed in the form \(\frac{p}{q}\), where \(p = 1\) and \(q = 280\). Since 1 and 280 are coprime, we can now find \(p + q\):

\[ p + q = 1 + 280 = 281 \]

### Final Answer
Thus, the final answer is:

\[ \boxed{281} \]