How many ways are there to choose six distinct integers from ${1, 2, . . . , 11}$ such that it is possible to attach each integer to a face of the cube where the sum of the integers on all pairs of opposite faces are equal?
The solution: We perform casework based on the sum of the opposite faces. The minimum possible sum is $1 + 6 = 7$ and the maximum possible sum is $6 + 11 = 17$. The number of corresponding face pair possibilities for the different sums are manually computed to be $3, 3, 4, 4, 5, 5, 5, 4, 4, 3, 3$ being symmetrical about where the sum equals 12. In total, there are $2 \cdot (2 \cdot $$3 \choose 3$$ + 2 \cdot $$4 \choose 3$$ + $$5 \choose 3$$) + $$5 \choose 3$$ = 50$ possibilities $\text{(D)}$.
However, I was pretty confused by what the solution meant when it referred to corresponding face pair possibilities. Are these different rotations of the same arrangement of six integers?
Anyone who could help me understand the solution and/or solve it in a cleaner way would be greatly appreciated.