We need to count the number of integer solutions to the equation:

$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 127$

where each $x_i$ satisfies $0 \leq x_i \leq 63$.

Step 1: Solve Without Constraints

Ignoring the upper bounds for now, the number of non-negative integer solutions to

$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 127$

is given by the stars and bars formula:

$\binom{127 + 6 - 1}{6 - 1} = \binom{132}{5}$

$= \frac{132!}{5!(127!)}$

$= \frac{132 \times 131 \times 130 \times 129 \times 128}{5!}$

$= \frac{132 \times 131 \times 130 \times 129 \times 128}{120}$

$= 96560646$

Step 2: Apply Inclusion-Exclusion for Upper Bound Constraint

Each variable $x_i$ must be at most 63. To correct for cases where some $x_i > 63$, we use the Inclusion-Exclusion Principle.

If we substitute $y_i = x_i - 64$ for some $x_i$, the new equation becomes:

$y_i + \sum_{j \neq i} x_j = 63$

which has solutions:

$\binom{68}{5} = \frac{68 \times 67 \times 66 \times 65 \times 64}{120} = 1426428.$

Since we can pick any one of the six variables to exceed 63, we subtract $6 \times 1426428$.

For overcorrection, if two variables exceed 63, then:

$y_1 + y_2 + \sum_{j \neq 1,2} x_j = -1,$

which has no solutions.

Thus, the final count is:

$96560646 - 6 \times 1426428 = 88001754.$

Final Answer:

$\boxed{88001754}$

Would you like a deeper explanation of any part? Here are some related problems to explore:

1. How does the stars and bars method work in combinatorics?
2. What is the Inclusion-Exclusion Principle, and how is it applied here?
3. How would the answer change if we had 7 variables instead of 6?
4. What if the sum was different, say 200 instead of 127?
5. Can this problem be solved using generating functions?

Tip: When dealing with bounded integer solutions, breaking it into unconstrained cases and then using Inclusion-Exclusion is a powerful method!