Question

Let P be the set of seven-digit numbers with the sum of their digits equal to 11. If the numbers in P are formed by using the digits 1, 2, and 3 only, then the number of elements in the set P is:

• A. 158
• B. 173
• C. 161
• D. 164

Hint:

Use the stars and bars method to count the number of solutions to equations involving non-negative integers.

Answer 1

The Correct Option is C

To ascertain the quantity of seven-digit numbers within set \( P \) composed exclusively of digits from \( \{1, 2, 3\} \) and summing to 11, we denote a number as \( d_1 d_2 d_3 d_4 d_5 d_6 d_7 \). The conditions are:

\( d_1 + d_2 + d_3 + d_4 + d_5 + d_6 + d_7 = 11 \)

Let \( n_1 \), \( n_2 \), and \( n_3 \) represent the frequencies of the digits 1, 2, and 3, respectively. The following constraints must be satisfied:

1. \( n_1 + n_2 + n_3 = 7 \) (total digit count),
2. \( 1 \cdot n_1 + 2 \cdot n_2 + 3 \cdot n_3 = 11 \) (digit sum),
3. \( n_1, n_2, n_3 \geq 0 \) and are integers.

1. Equation Simplification:
From the first constraint, \( n_1 = 7 - n_2 - n_3 \). Substituting this into the second constraint yields:

\( (7 - n_2 - n_3) + 2n_2 + 3n_3 = 11 \)

\( 7 + n_2 + 2n_3 = 11 \)

\( n_2 + 2n_3 = 4 \)

2. Non-Negative Integer Solutions:
Solving \( n_2 + 2n_3 = 4 \) for non-negative integers \( n_2 \) and \( n_3 \):

• Case 1: \( n_3 = 0 \). Then \( n_2 = 4 \). This implies \( n_1 = 7 - 4 - 0 = 3 \). The number of arrangements is \( \frac{7!}{3!4!0!} = 35 \).
• Case 2: \( n_3 = 1 \). Then \( n_2 = 4 - 2(1) = 2 \). This implies \( n_1 = 7 - 2 - 1 = 4 \). The number of arrangements is \( \frac{7!}{4!2!1!} = 105 \).
• Case 3: \( n_3 = 2 \). Then \( n_2 = 4 - 2(2) = 0 \). This implies \( n_1 = 7 - 0 - 2 = 5 \). The number of arrangements is \( \frac{7!}{5!0!2!} = 21 \).

For \( n_3>2 \), \( n_2 \) would be negative, rendering these solutions invalid.

3. Total Elements in Set \( P \):
The total count is the sum of arrangements from all valid cases:

\( 35 + 105 + 21 = 161 \).

Final Answer:
The total number of elements in set \( P \) is \( \boxed{161} \).