How many distinct ways are there to label the faces of a cube with distinct numbers from 1 to 6, such that there is at least one pair of opposite faces which do not sum to 7?

Rotations (which preserve orientation) are considered the same way.
Reflections (which do not preserve orientation) are considered distinct.

Without loss of generality, let the bottom face be one.

Then there are five choices for the top face and six choices for the other sides.
Of these, exactly one of them has faces which all sum to seven, hence the answer is $5 \cdot 6 - 1 = 29.$

To solve this problem, let's analyze the different cases based on the pairs of opposite faces:

1. Both opposite faces labeled with the numbers that sum up to 7:
In this case, we have only one way to label the cube. For example, if one opposite face is labeled with 1, the other opposite face must be labeled with 6.

2. One pair of opposite faces labeled with the numbers that sum up to 7:
There are 6 possible ways to choose one pair of opposite faces, and for each pair, we have 5 options to assign the numbers. This gives us a total of 6 * 5 = 30 ways.

3. Both pairs of opposite faces labeled with the numbers that do not sum up to 7:
In this case, there are no restrictions, and we have 6! = 720 ways to label the remaining faces.

Now, we can calculate the total number of distinct ways to label the faces of the cube:
Total = 1 (Case 1) + 30 (Case 2) + 720 (Case 3)
= 751

Therefore, there are 751 distinct ways to label the faces of a cube such that there is at least one pair of opposite faces which do not sum to 7.

To find the number of distinct ways to label the faces of a cube according to the given conditions, we can break down the problem into smaller steps and use basic counting principles.

Step 1: Count the total number of ways to label the faces of the cube without any restrictions.
Since there are 6 faces on a cube and we have 6 distinct numbers to label them with, there are 6! (6 factorial) ways to label the faces of the cube without any restrictions.

Step 2: Count the number of ways in which all opposite faces sum to 7.
There are three pairs of opposite faces on a cube: (1, 6), (2, 5), and (3, 4). For each pair, there are 2! (2 factorial) ways to assign the numbers. Therefore, there are (2!)^3 (2 factorial to the power of 3) ways to label the faces such that all opposite faces sum to 7.

Step 3: Count the number of cases where at least one pair of opposite faces do not sum to 7.
To find this, we subtract the number of cases from Step 2 from the total number of cases in Step 1:
Number of ways with at least one pair not summing to 7 = Total number of ways - Number of ways with all opposite faces summing to 7

Number of ways with at least one pair not summing to 7 = 6! - (2!)^3

Now, we can calculate the final answer: