JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Permutations and Combinations

A number is divisible by 11 if the difference between the sum of digits at odd positions and even positions is divisible by 11. Let the number be $abcdefg$. Then $(a+c+e+g)-(b+d+f)=11k$. Since $a+b+c+d+e+f+g=31$, we have either $(a+c+e+g)=21,(b+d+f)=10$ or $(a+c+e+g)=10,(b+d+f)=21$. Case 1: $(b,d,f)\in \{(1,2,7),(2,3,5),(1,4,5)\}$ (3 choices), $(a,c,e,g)$ are the remaining 4 digits (arranged in $4!$ ways). Total: $3\times 3!\times 4!=432$. Case 2: $(a,c,e,g)\in \{(1,2,3,4)\}$ (arranged in $4!$ ways), $(b,d,f)=(5,7,9)$ (arranged in $3!$ ways). Total: $4!\times 3!=144$. Grand total: $432+144=576$.