Key Concepts and Formulas

• Combinations with Constraints: Determining the number of ways to select digits that satisfy specific conditions, such as a target sum and a limited set of allowed digits.
• Permutations with Repetition: Calculating the number of distinct arrangements of n objects where some objects are identical. The formula is: $P = \frac{n!}{n_1! n_2! \ldots n_k!}$ where $n$ is the total number of objects, and $n_i$ is the number of objects of type i.
• Systematic Casework: Organizing the problem into mutually exclusive and exhaustive cases to ensure all possibilities are considered without overcounting.

Step-by-Step Solution

Step 1: Understanding the Problem and Setting up the Constraints

We need to find the number of 7-digit integers formed using only the digits 1, 2, and 3, such that the sum of the digits is equal to 10. Let $x$ be the number of 1s, $y$ the number of 2s, and $z$ the number of 3s. We have the following constraints:

$x + y + z = 7$ $x + 2y + 3z = 10$ $x, y, z \geq 0$ where $x, y, z$ are integers.

Step 2: Solving for Possible Combinations of x, y, and z

Subtract the first equation from the second equation to eliminate x: $(x + 2y + 3z) - (x + y + z) = 10 - 7$ $y + 2z = 3$ Now, we can analyze the possible non-negative integer values for z and find the corresponding y and x values:

• Case 1: z = 0 If $z = 0$, then $y + 2(0) = 3$, so $y = 3$. Substituting into $x + y + z = 7$, we get $x + 3 + 0 = 7$, so $x = 4$. Therefore, the combination is (4, 3, 0), meaning four 1s, three 2s, and zero 3s. The digit set is $\{1, 1, 1, 1, 2, 2, 2\}$.

• Case 2: z = 1 If $z = 1$, then $y + 2(1) = 3$, so $y = 1$. Substituting into $x + y + z = 7$, we get $x + 1 + 1 = 7$, so $x = 5$. Therefore, the combination is (5, 1, 1), meaning five 1s, one 2, and one 3. The digit set is $\{1, 1, 1, 1, 1, 2, 3\}$.

• Case 3: z = 2 If $z = 2$, then $y + 2(2) = 3$, so $y = -1$. This is not possible since $y \geq 0$.

Therefore, the only two possible combinations are (4, 3, 0) and (5, 1, 1).

Step 3: Calculating the Number of Arrangements for Each Combination

Now, we need to calculate the number of distinct 7-digit integers that can be formed from each combination.

• Combination (4, 3, 0): Four 1s, three 2s, zero 3s. Using the permutation formula: $P_1 = \frac{7!}{4!3!0!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1} = \frac{7 \times 6 \times 5}{6} = 35$

• Combination (5, 1, 1): Five 1s, one 2, one 3. Using the permutation formula: $P_2 = \frac{7!}{5!1!1!} = \frac{7 \times 6 \times 5!}{5!} = 7 \times 6 = 42$

Step 4: Summing the Arrangements to Find the Total Number of Integers

The total number of 7-digit integers is the sum of the arrangements from each valid combination: $Total = P_1 + P_2 = 35 + 42 = 77$

Common Mistakes & Tips

• Missing Cases: Ensure a systematic approach to finding all possible combinations to avoid missing valid scenarios.
• Incorrect Permutation Formula: Double-check the application of the permutation formula with repetitions. Ensure all factorials are correctly calculated.
• Arithmetic Errors: Be careful with arithmetic calculations, especially when simplifying factorial expressions.

Summary

We found two possible combinations of digits that sum to 10 using only 1, 2, and 3. These combinations were (4, 3, 0) and (5, 1, 1), representing the counts of 1s, 2s, and 3s, respectively. We then calculated the number of distinct permutations for each combination and summed them to find the total number of 7-digit integers.

The final answer is $\boxed{77}$, which corresponds to option (D).