Question

The digits of a nine-digit number are $1,2,3,4,5,6,7,8,9$, written in random order, the probability that the number is divisible by $11$ is

Answer 1

Let the 9 digit number be ABCDEFGHI

Since it is divisible by 11

s=(A+C+E+G+I)-(B+D+F+H) is divisible of 11

Sum of 1,2,3..9 is 45

S=45-2(B+D+F+H)

9<(B+D+F+H)<31

S is a odd number between -15,25

So only multiple of 11 in -15 to 25 are 11,-11

Total selection =${9_C}_4$=126

In this only 11 satisfies the solution

So probability=11/126