Suppose we have code with length of binary words 6. Like
000000
000001
But with distance 4 like
000000
001111
(meaning that between codewords at least 4 different bits) How many codewords in C(6,K,4)
I was able to brute-force that there is exactly 4 words.
But I have trouble to actually prove it - brute force is lame proof.
My first step is to prove that if 4 is minimal distance, there is no words with distance between them 5 or 6.
Simply put suppose we have 000000, and next word we can construct would be 001111 (no generality is lost there)
Then lets try to construct something of being as far as 5 bits from 000000 AND 001111.
Lets for beginning take first bits as 11: 11xxxx. 1 Seems natural choice since there is 0 in both words.
Then we have to choose 4 remaining bits to be at distance 2 from both 0000 and 1111, and that is just not possible.
If we try to put zero in first two bits, it also doesnt help to construct word with distance 5 from both first choosen words.
Thus if minimal distance of 6 bit codeword is 4, it is also maximal distance.
But how to count actual number of words in such code?
There are several ways. I'd go with a Hamming bound slightly improved.
If the code is linear, then there are $X=2^K$ codewords. Because the distance is 4, the set $S$ of all tuples, $|S|=2^N=64$, can be divided in the mutually exclusive sets:
$S_0:$ codewords ($|S_0| = X$)
$S_1:$ tuples at distance 1 from each codeword ($|S_1| = X \times N=6X$)
$S_2:$ tuples at distance 2 from codeword zero ($|S_2| = {N \choose 2} =15$)
$S_3:$ the rest
Then $X+6X+15\le 64\implies X\le 7$
Hence $X$ is (at most) $2^2=4$