JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Permutations and Combinations

A six-digit number is divisible by $11$ when the alternating sum $|(a+c+e)-(b+d+f)|$ is a multiple of $11$. The digits $0,1,2,5,7,9$ add to $24$, so if $a+c+e=S$ then $b+d+f=24-S$, and we need $|2S-24|$ to be $0$ or $11$; only $S=12$ works. One split summing to $12$ is $\{7,5,0\}$ and $\{9,2,1\}$: with $\{7,5,0\}$ in the odd places the leading digit cannot be $0$, giving $2\times 2!\times 3!=24$ numbers, while $\{9,2,1\}$ in the odd places has no zero to worry about, giving $3!\times 3!=36$. Altogether $24+36=60$.