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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

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Solution

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 =9C4=126
In this only 11 satisfies the solution
So probability=11/126















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