Placement with repetitions?

Question

Combinatorics

What is the maximum number of binary words of length 10 (sequences of ten zeros and ones) that any two chosen words differ in at least two places?

There are 2^10 combinations in total, but how to choose words that differ in at least two places?

Answer 1

half of all, that is, 2^9 = 512
choose zero, all numbers with 2 ones, all with 4 ones, etc., up to 1111111111. It is easy to calculate that their number is 2^9
1 + C(10, 2) + C(10, 4) + C(10, 6) + C(10, 8) + 1 = 512
Why not more? With each choice of one word, we cross out 10 neighboring ones (which differ from the selected one by one bit). According to this algorithm, each word can be crossed out no more than 10 times. If we choose more than 512, then there will be more than 5120 individual crossed out words, and less than 512 crossed out words (total 1024), that is, someone was crossed out more than 10 times, a contradiction.