JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Permutations and Combinations

The digits $1,2,2,3$ give $\frac{4!}{2!}=12$ distinct four-digit numbers. Fixing $1$ in a chosen place leaves $2,2,3$ to fill the rest in $\frac{3!}{2!}=3$ ways, and the same count holds for $3$, so each of $1$ and $3$ stands in any given place $3$ times and $2$ stands there $12-3-3=6$ times. The digits contributed to each place therefore add to

and since every place carries the same total, the sum of all twelve numbers is $24\times 1111=26664$.