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 solve the problem, we need to determine the number of seven-digit numbers in set \( P \) where each digit is from the set \( \{1, 2, 3\} \), and the sum of the digits equals 11. Let \( d_1 d_2 d_3 d_4 d_5 d_6 d_7 \) represent the seven-digit number, with the following conditions:

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

Let \( n_1 \), \( n_2 \), and \( n_3 \) represent the number of times the digits 1, 2, and 3 appear, respectively. The following constraints apply:

1. \( n_1 + n_2 + n_3 = 7 \) (total number of digits is 7),
2. \( 1 \cdot n_1 + 2 \cdot n_2 + 3 \cdot n_3 = 11 \) (sum of the digits is 11),
3. \( n_1, n_2, n_3 \geq 0 \) and are integers.

1. Simplifying the Equations:
From the first equation, \( n_1 = 7 - n_2 - n_3 \). Substituting this into the second equation:

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

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

\( n_2 + 2n_3 = 4 \)

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

• If \( n_3 = 0 \), then \( n_2 = 4 \). Substituting into \( n_1 = 7 - n_2 - n_3 \):
  
  \( n_1 = 7 - 4 - 0 = 3 \).
  The number of arrangements is:
  
  \( \frac{7!}{3!4!0!} = \frac{5040}{6 \times 24 \times 1} = \frac{5040}{144} = 35 \).
• If \( n_3 = 1 \), then \( n_2 = 4 - 2(1) = 2 \). Substituting into \( n_1 = 7 - n_2 - n_3 \):
  
  \( n_1 = 7 - 2 - 1 = 4 \).
  The number of arrangements is:
  
  \( \frac{7!}{4!2!1!} = \frac{5040}{24 \times 2 \times 1} = \frac{5040}{48} = 105 \).
• If \( n_3 = 2 \), then \( n_2 = 4 - 2(2) = 0 \). Substituting into \( n_1 = 7 - n_2 - n_3 \):
  
  \( n_1 = 7 - 0 - 2 = 5 \).
  The number of arrangements is:
  
  \( \frac{7!}{5!0!2!} = \frac{5040}{120 \times 1 \times 2} = \frac{5040}{240} = 21 \).

If \( n_3 > 2 \), then \( 2n_3 > 4 \), which would make \( n_2 \) negative. Thus, there are no other valid solutions.

3. Total Number of Elements in Set \( P \):
The total number of elements in set \( P \) is the sum of the number of arrangements for all cases:

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

Final Answer:
The final answer is \( \boxed{161} \).