Note that each group of numbers should sum to 31+2+3+4+5+6+7+8+9=15. Thus, this is equivalent to asking, "How many ways can you fill in a three by three magic square with the integers 1 through 9?" since we can take the three rows of the magic square as our three groups. If you have closely studied magic squares, you might know that in a three by three magic square that is to be filled in with the integers 1 through 9, the center of the square would be 5 (the average of the numbers), and the numbers in the corners should be even(*). The such pairs (disregarding order) are (2,8) and (4,6). Let's fix the position of 2 to be the top left corner. This would make 8 in the bottom right corner. We can have either 4 or 6 to be in the top right corner, for a total of 2 such groups of three. (The groups are (8,3,4) (1,5,7) (6,9,2) and (8,1,6) (3,5,7) (4,9,2). ) Note that if we had instead fixed the position of 4, 6, or 8, they would correspond to one of the two cases, just in a different configuration. (*)We can prove this using proof by contradiction. Label the nine small squares within the magic square from a to i from left to right, top to bottom. Firstly, we know that a+i and c+g sum to 10 since the center square is 5. Thus, a and i must have the same parity, and so must c and g. Suppose that a and c have different parity. Since a+b+c=15, b must be even. By a similar argument, h must also be even, and so must d and f. Our initial assumption is that one of a and c is odd and the other is even; however, we end up with six even numbers needed to fill in the square, but there are only four even integers from 1 to 9. Now suppose that all of $a, c, g,$ and i are odd. This would make each of $b, d, f,$ and h odd, but clearly we do not have enough odd numbers to make all nine numbers odd. Thus, each corner square must be even. ~ Brian__Liu To start, we must find the value that all three groups must add to, which is $\frac{1+2+3+4+5+6+7+8+9}{3} = 15.$. Then, we can notice that the numbers 1, 2, and 3 must be in different groups because two of them and any other digit 1-9 added together cannot produce 15 (For example, $2+3=5$ which means that we need a 10 to sum to 15. In the same way, 7, 8, and 9 cannot be placed in the same group because putting them in the same group would make the numbers greater than 15 (For example, $7+8=15$, meaning that it would be impossible to add another number with making the sum greater than 15. Now, we can just use trial and error to solve the problem (this is feasible because there are only $3*2=6$ ways to pair a number $1-3$ and $7-9$ and then add another number $4-6$ in order to sum to 15. By trial and error, we see that 1 and 7 must not be paired together because to have a sum of 15, there must be another 7 in the group (The problem only lets us use 1 of each number). This eliminates two of the options we are considering. Now, we only have to check the 4 options where 7 is not in a group with 1. By trial and error, we can see that only two cases work: 1. $(1,8,6); (2,9,4); (3,7,5)$ and 2. $(1,9,5); (2,7,6); (3,8,4)$ Because we only have two options, the answer is $C$ (Too bashy, not recommended in the real test if you encounter these problems)