Suppose we have $6$ symbols, say $A,B,C,D,E,F$. We are asked to form $4$ new symbols using the $6$ symbols with the addition operation.
For example, the $4$ new symbols can be $A+C+E, F+E+A, D+C+A, A+F+D$. There are some restrictions on the symbols, they are
$1)$each symbol appears only once in the $4$ new symbols
$2)$ the total number of each symbols in the $4$ new symbols must be less than $4$
I am looking for a way to make the $4$ symbols contain as many common symbols as possible. Can anyone help me?
I do not write all $+$ out for better visibility.
I think the way of maximizing the number of common symbols results in the following (or similar by permutation) solution
$$ABCD,ABCD+E,ABCD+F,ABCD+EF$$ which has 4 common symbols, which is maximal. You cannot have 6 commons symbols because this will yield only 1 new symbol, and you cannot have 5 common symbols because this will yield only two new symbols. ($ABCDE,ABCDE+F$).