Eliana has a box of tiles, each with an integer from \(0\) to \(9\) on it. Each integer appears on at least three tiles. Eliana creates larger numbers by placing tiles side by side. For example, using the tiles \(3\) and \(7\), Eliana can create the \(2\)digit number \(37\) or \(73\).
Using six of her tiles, Eliana forms two \(3\)digit numbers, \(ABC\) and \(DEF\), that add to \(1234\).
Eliana then notices that \(A>D,~B>E,\) and \(C>F\). How many possible \(6\)tuples \((A,~B,~C,~D,~E,~F)\) could she have chosen?
Since \(C + F\) ends in a \(4\), then \(C+F = 4\) or \(C+F = 14\). The value of \(C+F\) cannot be \(20\) or more, because \(C\) and \(F\) are digits. In the case that \(C+F= 14\), we “carry” a \(1\) to the tens column. Now we will look at the tens column for these two cases.
Case 1: \(C+F = 4\)
Since the result in the tens column is \(3\) and there was no “carry” from the units column, it follows that \(B+E\) ends in a \(3\). Then \(B+E=3\) or \(B+E=13\). The value of \(B+E\) cannot be \(20\) or more, because \(B\) and \(E\) are digits. In the case that \(B+E = 13\), we “carry” a \(1\) to the hundreds column.
Case 2: \(C+F=14\)
Since the result in the tens column is \(3\) and there was a “carry” from the units column, it follows that \(1+B+E\) ends in a \(3\), so \(B+E\) ends in a \(2\). Then \(B+E=2\) or \(B+E=12\). The value of \(B+E\) cannot be \(20\) or more, because \(B\) and \(E\) are digits. In the case that \(B+E = 12\), we “carry” a \(1\) to the hundreds column.
Since the result in the hundreds column is \(12\), then \(A+D = 12\), or in the case when there was a “carry” from the tens column, \(1 + A + D =12\), so \(A+D = 11\).
We summarize this information in the following tree.
We now look at the different possibilities for digits \(A, ~B, ~C, ~D, ~E,\) and \(F\) for each individual sum, with the restriction that \(A>D\), \(B>E\), and \(C>F\).
Sum  Solutions  Number of Solutions 

\(C+F=4\) 

\(2\) 
\(C+F=14\) 

\(2\) 
\(B+E=2\) 

\(1\) 
\(B+E=3\) 

\(2\) 
\(B+E=12\) 

\(3\) 
\(B+E=13\) 

\(3\) 
\(A+D=11\) 

\(4\) 
\(A+D=12\) 

\(3\) 
Since we can use each number at least 3 times, all combinations of solutions outlined in the table are possible, and so we can then update our tree diagram with the total number of solutions for each sum.
The first (leftmost) path through the tree corresponds to the sums \(C+F=4\) (2 solutions), \(B+E=3\) (2 solutions), and \(A+D=12\) (3 solutions). Since there are 2 ways to achieve the first sum, and for each of these possibilities there are 2 ways to achieve the second sum, and for each of these possibilities there are 3 ways to achieve the third sum, the number of 6tuples this path corresponds to is equal to \(2 \times 2 \times 3 = 12\).
Similarly, the second path through the tree corresponds to the sums \(C+F=4\) (2 solutions), \(B+E=13\) (3 solutions), and \(A+D=11\) (4 solutions). So the number of 6tuples this path corresponds to is equal to \(2 \times 3 \times 4= 24\).
The third path through the tree corresponds to the sums \(C+F=14\) (2 solutions), \(B+E=2\) (1 solution), and \(A+D=12\) (3 solutions). So the number of 6tuples this path corresponds to is equal to \(2 \times 1 \times 3= 6\).
The fourth path through the tree corresponds to the sums \(C+F=14\) (2 solutions), \(B+E=12\) (3 solutions), and \(A+D=11\) (4 solutions). So the number of 6tuples this path corresponds to is equal to \(2 \times 3 \times 4= 24\).
Therefore, the total number of 6tuples is \(12+24+6+24=66\).