Question

an eight digit number is formedby 1 to 8digits without repetition. how many number formed thus divisible by 11.

Answer 1

The sum of all ten digits is 4545, which is divisible by 99 (so any number formed using these digits will be divisible by nine). But we need to form an eight digit number, so we must remove two digits. In order to have the remaining number of digits still add up to a multiple of 99, the two digits removed must also add up to a multiple of 99. In fact, they must add up to exactly 99.

Their are five such pairs: (0,9),(1,8),…,(4,5)(0,9),(1,8),…,(4,5). How many eight digit numbers can be formed with the digits left after removing one such pair?

If we remove 00 and 99, we can form an eight digit number with the remaining digits in 8!8! ways.

If we remove any other pair, then 00 is one of the remaining digits, and we must take care that the number formed must not have a leading zero (for otherwise it would be a seven digit number). There are seven choices for the first position if we exclude 00, and then the remanining 77 digits can be arranged in 7!7! ways. So there are 7×7!7×7!numbers of this kind, corresponding to each of the four pairs (1,8),…,(4,5)(1,8),…,(4,5).

All cases above are distinct (for in each case, the number contains some digits that are always missing in the other cases). So the total number of such eight digit numbers is 8!+4×7×7!=36×7!8!+4×7×7!=36×7!, or 181440181440.
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Answer 2

12345678
23456781
34567812
45678123
56781234
67812345
78123456
81234567