JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Permutations and Combinations

A $3\times 3$ matrix has 9 entries, each from $\{-1,0,1\}$. We need the sum of all entries to equal 5.

Let $x$ be the number of 1's, $y$ be the number of $-1$'s, and $z$ be the number of 0's.

Constraints: $x+y+z=9$ and $x-y=5$.

From $x-y=5$, we get $x=y+5$.

Substituting: $(y+5)+y+z=9$, so $2y+z=4$.

Case I: $y=2,z=0$: $x=7$. Number of arrangements: $\frac{9!}{7!2!}=36$.

Case II: $y=1,z=2$: $x=6$. Number of arrangements: $\frac{9!}{6!1!2!}=252$.

Case III: $y=0,z=4$: $x=5$. Number of arrangements: $\frac{9!}{5!4!}=126$.

Total: $36+252+126=414$.