JEE Main 2023 — Mathematics Algebra

Given,

A $5\times 5$5×5 matrix whose each entry is either $0$0 or $1$1, is such that sum of entries of each column as well as each row is $1$1,

Taking a possible case we get,

$$[\begin{matrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1\end{matrix}]$$[10000​01000​00100​00010​00001​]

Now arranging the above case we get,

[\begin{matrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1\end{matrix}]\begin{matrix}\rightarrow 5\text{ways} \\ \rightarrow 4\text{ways} \\ \rightarrow 3\text{ways} \\ \rightarrow 2\text{ways} \\ \rightarrow 1\text{way}\end{matrix}} because if we fix $1$1 at first row and first column then in second row first column $1$1 cannot come and vice-versa.

So total number of such matrices will be $5\times 4\times 3\times 2\times 1=120$5×4×3×2×1=120