JEE Mains 24-Jan-2023 Shift 2 Solved Paper

In each row and each column exactly one is to be placed -
∴ No. of such matrices =5×4×3×2×1=120
Alternate :Step-1 : Select any 1 place for 1 's in row 1. Automatically some column will get filled with 0 's.
Step-2 : From next now select 1 place for 1 's. Automatically some column will get filled with 0 's. ⇒ Each time one less place will be available for putting 1's.
Repeat step-2 till last row.
Req. ways =5×4×3×2×1=120