Let A be the set of all `3xx3` symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them of 0.
The number of matrices in A is

Options

A. 12
B. 6
C. 9
D. 3

Answer

To solve the problem, we need to find the number of symmetric \(3 \times 3\) matrices with entries either 0 or 1, where exactly 5 entries are 1 and 4 entries are 0.

### Step-by-Step Solution:

1. **Understanding the Structure of a Symmetric Matrix:**
A symmetric \(3 \times 3\) matrix has the form:
\[ \begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \end{pmatrix} \]
Here, the entries \(a, b, c, d, e, f\) are the only independent entries since the matrix is symmetric.

2. **Counting the Independent Entries:**
In total, there are 6 independent entries in a \(3 \times 3\) symmetric matrix: \(a, b, c, d, e, f\).

3. **Setting Up the Problem:**
We need to choose 5 entries to be 1 and 1 entry to be 0. Since there are 6 independent entries, we can choose which of these will be 0.

4. **Choosing the Entry to be 0:**
We can choose 1 entry to be 0 from the 6 independent entries. The number of ways to choose 1 entry from 6 is given by:
\[ \binom{6}{1} = 6 \]

5. **Filling the Remaining Entries with 1:**
After choosing 1 entry to be 0, the remaining 5 entries will automatically be filled with 1s.

6. **Conclusion:**
Therefore, the total number of symmetric \(3 \times 3\) matrices with 5 entries as 1 and 1 entry as 0 is:
\[ \text{Total Matrices} = 6 \]

### Final Answer:
The number of matrices in \(A\) is \(6\).