How many number can be formed with 6 digit where the number formed by the 1st,2nd and 4th digit(count should be from left side) and the number formed by the 3rd,5th and 6th digit also should be divisible by 11?
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there is no intersection of digits between the first and second number formed. Thus the two numbers can be any 3-digit multiple of 11. There are 81 3-digit numbers which are multiples of 11 (take the floor of 999/11 and subtract the floor of 99/11). Thus the total combinations is 81^2=6561