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(B) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$,where $a, b, c, d \in \{0, 1, 2, 3, 4, 5\}$. The sum $S = a + b + c + d = p$,where $p \in \

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$180$

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(B) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$,where $a, b, c, d \in \{0, 1, 2, 3, 4, 5\}$. The sum $S = a + b + c + d = p$,where $p \in \{3, 5, 7\}$.Case $(i): S = 3$. The number of non-negative integer solutions to $a + b + c + d = 3$ is given by $\binom{3+4-1}{4-1} = \binom{6}{3} = 20$.Case $(ii): S = 5$. The number of non-negative integer solutions to $a + b + c + d = 5$ is given by $\binom{5+4-1}{4-1} = \binom{8}{3} = 56$.Case $(iii): S = 7$. The number of non-negative integer solutions to $a + b + c + d = 7$ with $a, b, c, d \le 5$ is calculated using inclusion-exclusion. Total solutions without restriction is $\binom{7+4-1}{4-1} = \binom{10}{3} = 120$. Subtract cases where at least one variable $\ge 6$. If $a \ge 6$,let $a = a' + 6$,then $a' + b + c + d = 1$,which has $\binom{1+4-1}{4-1} = \binom{4}{3} = 4$ solutions. Since there are $4$ variables,total invalid cases = $4 	imes 4 = 16$. Thus,$120 - 16 = 104$.Total number of matrices = $20 + 56 + 104 = 180$.

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 $p$,where $2 < p < 8$,then the number of such matrices $A$ is:

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