Associate each letter with the correct number, given that:
- If A is 1, then B is not 3.
- If B is not 1, then D is 4.
- If B is 1, then C is 4.
- If C is 3, then D is not 2.
- If C is not 2, then D is 2.
- If D is 3, then A is not 4.
[BDAC] = 
"Associate each letter with the correct Distinct number "
makes more sense and challenge' than without"Distinct "
If such is the text (my assumption ), then statement 3 & statement 5 force the BDAC as the only possible soution.
We verify that this solution causes no contradiction vs other statements.
IT DOES NOT.