Question

The number of $3\times 3$ matrices that can be formed using the elements $0,1,2,3$ such that all principal diagonal elements are different, is

• A. $6\times 4^6$
• B. $6\times 4^7$
• C. $6\times 3^7$
• D. $4^7$

Answer 1

The correct option is B $6\times 4^7$

$\left(a\right)$ Since, diagonal elements are different, number of ways of arranging them $=4\times 3\times 2$
$\left(b\right)$ Number of ways of arranging other ($6$ remaining) elements $=4^6$
Hence, total number of matrices $=4\times 3\times 2\times 4^6=6\times 4^7$