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

Options

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

Explanation

To solve the problem of finding the number of `3 x 3` symmetric matrices with entries either 0 or 1, where five entries are 1 and four entries are 0, we can follow these steps:

### Step 1: Understand the Structure of a Symmetric Matrix
A `3 x 3` symmetric matrix has the following structure:
\[ \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{pmatrix} \]
This means that the elements above the diagonal mirror those below the diagonal, and the diagonal elements can be chosen independently.

### Step 2: Count the Total Number of Entries
In a `3 x 3` matrix, there are 9 entries. For a symmetric matrix, we can represent it with 6 independent entries:
- 3 diagonal entries: \(a_{11}, a_{22}, a_{33}\)
- 3 off-diagonal entries: \(a_{12}, a_{13}, a_{23}\)

### Step 3: Set the Conditions
We need to use exactly 5 entries as 1 and 4 entries as 0. This means we need to consider the placement of 1s in the 6 independent positions.

### Step 4: Case Analysis
We will analyze different cases based on the number of diagonal entries that are 1.

#### Case 1: All diagonal entries are 1
- Diagonal: \(1, 1, 1\) (3 ones used)
- Off-diagonal: We need 2 more 1s from \(a_{12}, a_{13}, a_{23}\). We can select any 2 from these 3 positions.
- The number of ways to choose 2 positions from 3 is given by \(\binom{3}{2} = 3\).

#### Case 2: Two diagonal entries are 1, one is 0
- Diagonal: \(1, 1, 0\) (2 ones used)
- Off-diagonal: We need 3 more 1s. The only way to do this is to set all off-diagonal entries to 1.
- The arrangement is \(1, 1, 0\) on the diagonal and \(1, 1, 1\) off-diagonal.
- The number of ways to arrange the diagonal is given by \(\binom{3}{2} = 3\) (choosing 2 positions for 1s out of 3).

#### Case 3: One diagonal entry is 1, two are 0
- Diagonal: \(1, 0, 0\) (1 one used)
- Off-diagonal: We need 4 more 1s, but this is impossible since we only have 3 off-diagonal positions. Thus, this case is not possible.

#### Case 4: All diagonal entries are 0
- Diagonal: \(0, 0, 0\) (0 ones used)
- Off-diagonal: We need 5 ones, but we can only place 3 ones in the off-diagonal positions. Thus, this case is also not possible.

### Step 5: Total Count of Matrices
From the valid cases:
- Case 1 contributes 3 matrices.
- Case 2 contributes 3 matrices.

Thus, the total number of symmetric matrices is:
\[ 3 + 3 = 6 \]

### Final Answer
The total number of symmetric matrices in set A is **6**.