• # question_answer The number of 6 digit numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digit is repeated, is : [JEE Main 10-4-2019 Morning] A) 36                               B) 60C) 48                               D) 72

Sum of given digits 0, 1, 2, 5, 7, 9 is 24. Let the six digit number be abcdef and to be divisible by 11 so $|(a+c+e)-(b+d+f)|$is multiple of 11. Hence only possibility is $a+c+e=12=b+d+f$ Case-I $\left\{ a,c,e \right\}=\left\{ 9,2,1 \right\}\And \left\{ b,d,f \right\}=$$\left\{ 7,5,0 \right\}$ So, Number of numbers $=3!\times 3!=36$ Case-II $\left\{ a,c,e \right\}=\left\{ 7,5,0 \right\}$and $\left\{ b,d,f \right\}=\left\{ 9,2,1 \right\}$ So, Number of numbers $2\times 2!\times 3!=24$ Total = 60