Let A be a matrix of order 2 × 2 , whose entries are from the set \...

Given, A = \left[\begin{cases}a & b \\ c & d\end{cases}\right⌋ and $a, b, c, d \in \left\{0, 1, 2, 3, 4, 5\right\}$

Let, $a + b + c + d = p, p \in \left\{3, 5, 7\right\}$

Case-i

$a + b + c + d = 3$ where $a, b, c, d \in \left\{0, 1, 2, 3\right\}$

Entries will be \left(0, 0, 0, 3\right)\text{or}\left(0, 0, 1, 2\right)\text{or}\left(0, 1, 1, 1\right)$</p><p>Number of ways$= \dfrac{4 !}{3 !} + \dfrac{4 !}{2 !} + \dfrac{4 !}{3 !} = 20 ⋯ \left(1\right)$</p><p>Case-ii</p><p>$a + b + c + d = 5$where$a, b, c, d \in \left\{0, 1, 2, 3, 4, 5\right\}$</p><p>Entries will be \left(0, 0, 0, 5\right)\text{or}\left(0, 0, 1, 4\right)\text{or}\left(0, 0, 2, 3\right)\text{or}\left(0, 3, 1, 1\right)\text{or}\left(0, 1, 2, 2\right)\text{or}\left(1, 1, 1, 2\right)$

Number of ways $= \dfrac{4 !}{3 !} + \dfrac{4 !}{2 !} + \dfrac{4 !}{2 !} + \dfrac{4 !}{2 !} + \dfrac{4 !}{2 !} + \dfrac{4 !}{3 !} = 56 ⋯ \left(2\right)$

Case-iii

$a + b + c + d = 7$

Number of ways = total number of ways when $a, b, c, d \in \left\{0, 1, 2, 3, 4, 5, 6, 7\right\}$-total number of ways when $a, b, c, d \notin \left\{6, 7\right\}$

Number of ways ${=}^{7 + 4 - 1}{C}_{4 - 1} - \left(\dfrac{4 !}{3 !} + \dfrac{4 !}{2 !}\right)$

$= C3 10 - 16 = 104 ⋯ \left(3\right)$

Hence adding all equation so total number of ways $= 180$