Let A be a matrix of order 2 × 2 , whose entries are from the set 0 , 1 , 2 , 3 , 4 , 5 . If the sum of…

Given, $A=\left[\begin{array}{ll}a& b\\ c& d\end{array}\right.⌋$ 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 $\left(0,0,0,3\right) \text{or} \left(0,0,1,2\right) \text{or} \left(0,1,1,1\right)$

Number of ways $=\frac{4!}{3!}+\frac{4!}{2!}+\frac{4!}{3!}=20 \cdots \left(1\right)$

Case-ii

$a+b+c+d=5$ where $a,b,c,d\in \left\{0,1,2,3,4,5\right\}$

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 $=\frac{4!}{3!}+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{3!}=56 \cdots \left(2\right)$

Case-iii

$a+b+c+d=7$

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

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

$={}^{10}C_3-16=104 \cdots \left(3\right)$

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