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 placing 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 can follow these steps:

### Step 1: Calculate the Total Number of Arrangements
The total number of arrangements of the letters (3 A's, 3 B's, and 3 C's) in the 3x3 matrix can be calculated using the formula for permutations of multiset:

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

Where:
- \(9!\) is the factorial of the total number of letters.
- \(3!\) is the factorial for each of the identical letters (A's, B's, and C's).

Calculating this gives:

\[ 9! = 362880 \]
\[ 3! = 6 \]
\[ \text{Total arrangements} = \frac{362880}{6 \times 6 \times 6} = \frac{362880}{216} = 1680 \]

### Step 2: Calculate the Favorable Arrangements
Next, we need to find the number of favorable arrangements where no row or column contains two identical letters. This can be visualized as arranging the letters in a way that each letter appears exactly once in each row and column.

This is equivalent to finding the number of ways to arrange three distinct letters (A, B, C) in a 3x3 grid such that each letter appears exactly once in each row and column. This is a classic problem of arranging distinct objects in a square grid, which can be solved using the concept of permutations.

The number of favorable arrangements can be calculated as follows:

1. Choose a permutation of the letters for the first row (3! ways).
2. For the second row, we can choose any permutation of the remaining letters that does not repeat any letter in the same column (2 ways).
3. For the third row, there is only 1 way to arrange the remaining letters.

Thus, the number of favorable arrangements is:

\[ \text{Favorable arrangements} = 3! \times 2 \times 1 = 6 \times 2 \times 1 = 12 \]

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

\[ \text{Probability} = \frac{\text{Favorable arrangements}}{\text{Total arrangements}} = \frac{12}{1680} \]

Simplifying this fraction:

\[ \frac{12}{1680} = \frac{1}{140} \]

### Step 4: Express in Terms of Coprime p and q
The probability can be expressed as \(\frac{p}{q}\) where \(p = 1\) and \(q = 140\). Since 1 and 140 are coprime, we can find \(p + q\):

\[ p + q = 1 + 140 = 141 \]

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

\[ \boxed{141} \]