Key Concepts and Formulas

• Generating Functions: A power series where the coefficient of $x^k$ represents the number of ways to achieve a sum of $k$.
• Geometric Series Sum: $1 + r + r^2 + ... + r^{n-1} = \frac{1 - r^n}{1 - r}$.
• Binomial Theorem (Generalized): $(1 + x)^n = \sum_{r=0}^{\infty} \binom{n}{r} x^r$, where $\binom{n}{r} = \frac{n(n-1)...(n-r+1)}{r!}$ for any real number $n$. If $n$ is a negative integer, say $n=-k$, then $\binom{-k}{r} = (-1)^r \binom{k+r-1}{r}$.

Step-by-Step Solution

Step 1: Define the Problem Mathematically

We need to find the number of solutions to the equation: $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 12$ subject to the constraint $1 \le x_i \le 4$ for all $i = 1, 2, ..., 7$. Each $x_i$ represents a digit in our seven-digit number.

Step 2: Construct the Generating Function

Since each digit $x_i$ can take values 1, 2, 3, or 4, the generating function for a single digit is: $(x^1 + x^2 + x^3 + x^4)$ For seven digits, we raise this to the power of 7: $G(x) = (x^1 + x^2 + x^3 + x^4)^7$ We want to find the coefficient of $x^{12}$ in the expansion of $G(x)$.

Step 3: Simplify the Generating Function

Factor out $x$ from the parenthesis: $G(x) = (x(1 + x + x^2 + x^3))^7 = x^7(1 + x + x^2 + x^3)^7$ To find the coefficient of $x^{12}$ in $G(x)$, we need to find the coefficient of $x^{12-7} = x^5$ in $(1 + x + x^2 + x^3)^7$.

Use the formula for the sum of a geometric series: $1 + x + x^2 + x^3 = \frac{1 - x^4}{1 - x}$. Substituting this into the expression: $G(x) = x^7 \left( \frac{1 - x^4}{1 - x} \right)^7 = x^7 (1 - x^4)^7 (1 - x)^{-7}$ Now we are looking for the coefficient of $x^5$ in $(1 - x^4)^7 (1 - x)^{-7}$.

Step 4: Expand Using the Binomial Theorem

Expand $(1 - x^4)^7$ using the binomial theorem: $(1 - x^4)^7 = \binom{7}{0} - \binom{7}{1}x^4 + \binom{7}{2}x^8 - \binom{7}{3}x^{12} + ... = 1 - 7x^4 + 21x^8 - 35x^{12} + ...$

Expand $(1 - x)^{-7}$ using the generalized binomial theorem: $(1 - x)^{-7} = \sum_{r=0}^{\infty} \binom{-7}{r} (-x)^r = \sum_{r=0}^{\infty} \binom{7+r-1}{r} x^r = \sum_{r=0}^{\infty} \binom{6+r}{r} x^r$ Writing out the first few terms: $(1 - x)^{-7} = \binom{6}{0} + \binom{7}{1}x + \binom{8}{2}x^2 + \binom{9}{3}x^3 + \binom{10}{4}x^4 + \binom{11}{5}x^5 + ...$

Step 5: Find the Coefficient of $x^5$

We need the coefficient of $x^5$ in the product $(1 - 7x^4 + ...)(\binom{6}{0} + \binom{7}{1}x + \binom{8}{2}x^2 + \binom{9}{3}x^3 + \binom{10}{4}x^4 + \binom{11}{5}x^5 + ...)$. The terms that contribute to $x^5$ are: \begin{itemize} \item $1 \cdot \binom{11}{5}x^5 = \binom{11}{5}x^5$ \item $-7x^4 \cdot \binom{7}{1}x = -7\binom{7}{1}x^5$ \end{itemize} Thus, the coefficient of $x^5$ is $\binom{11}{5} - 7\binom{7}{1}$.

Step 6: Calculate the Binomial Coefficients

$\binom{11}{5} = \frac{11!}{5!6!} = \frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 11 \cdot 3 \cdot 2 \cdot 7 = 462$ $\binom{7}{1} = \frac{7!}{1!6!} = 7$ Therefore, the coefficient of $x^5$ is $462 - 7 \cdot 7 = 462 - 49 = 413$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when using the generalized binomial theorem, especially with negative exponents.
• Index Calculation: Double-check the indices in the binomial coefficients to ensure you are selecting the correct terms for multiplication.
• Over-Expansion: Avoid expanding the series too far. Only calculate terms that are necessary to find the coefficient of the target power of $x$.

Summary

We used generating functions to solve this combinatorial problem. By constructing the generating function, simplifying it using geometric series and binomial theorem, and extracting the relevant coefficient, we found the number of seven-digit integers satisfying the given conditions. The number of such integers is 413.

Final Answer

The final answer is \boxed{413}.