The number of 3×3 non-singular matrices with four entires as 1 and all other entries as 0 is

$\text{ Let }A=\left[\begin{array}{l}{a}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}_3\\ b_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{b}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{b}_3\\ c_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{c}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{c}_3\end{array}\right]$
$\therefore \det .\left(A\right)=a_1{b}_2{c}_3+a_2{b}_3{c}_1+b_1{c}_2{a}_3-a_3{b}_2{c}_1-a_2{b}_1{c}_3-a_1{b}_3{c}_2$
Hence number of non–singular matrices
= (no. of ways of choosing any one triplet) × (no. of ways of choosing any one element) ${=}^6{C}_1{\times }^6{C}_1$