JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Permutations and Combinations

All four digits are different with largest digit $7$, so they are drawn from $\{0,1,\dots ,7\}$ with $7$ present, and the four digits split into a front pair and a back pair of equal sum $s$. One pair must contain $7$, namely $\{7,s-7\}$, and the other is a disjoint pair of distinct digits also summing to $s$. The admissible choices are, by $s$: for $s=7$, $\{1,6\},\{2,5\},\{3,4\}$ together with $\{7,0\}$; for $s=8$, $\{2,6\},\{3,5\}$ with $\{7,1\}$; for $s=9$, $\{3,6\},\{4,5\}$ with $\{7,2\}$; for $s=10$, $\{4,6\}$ with $\{7,3\}$; for $s=11$, $\{5,6\}$ with $\{7,4\}$ — nine choices in all. Each choice fixes the two pairs, and the code is then formed by selecting the front pair ($2$ ways) and ordering each pair ($2\times 2$), giving $8$ codes per choice. Hence the number of possible codes is $9\times 8=72$, which is the maximum number of trials required.