Let $$A$$ be a matrix of order $$2 \times 2$$, whose entries are from the set $$\{0, 1, 2, 3, 4, 5\}$$. If the sum of all the entries of $$A$$ is a prime number $$p, 2 < p < 8$$, then the number of such matrices $$A$$ is ______

We need to count $$2 \times 2$$ matrices with entries from $$\{0, 1, 2, 3, 4, 5\}$$ whose entry sum is a prime $$p$$ with $$2 < p < 8$$.

The primes in the range $$(2, 8)$$ are $$p = 3, 5, 7$$.

We need the number of solutions to $$a + b + c + d = p$$ where $$0 \leq a, b, c, d \leq 5$$.

Case 1: $$p = 3$$

Without the upper bound constraint, the number of non-negative integer solutions is $$\binom{3+3}{3} = \binom{6}{3} = 20$$.

Since no variable can exceed 3 when the sum is 3, all solutions satisfy the upper bound. Count = 20.

Case 2: $$p = 5$$

Unrestricted: $$\binom{8}{3} = 56$$.

Since the sum is 5, no single variable can exceed 5, so the upper bound constraint is automatically satisfied. Count = 56.

Case 3: $$p = 7$$

Unrestricted: $$\binom{10}{3} = 120$$.

Subtract cases where any variable $$\geq 6$$: if $$a \geq 6$$, let $$a' = a - 6$$, then $$a' + b + c + d = 1$$, giving $$\binom{4}{3} = 4$$ solutions.

By symmetry across all 4 variables: subtract $$4 \times 4 = 16$$.

Count = $$120 - 16 = 104$$.

Total:

$$20 + 56 + 104 = 180$$

The correct answer is $$\boxed{180}$$.