How many 5 digit numbers contain exactly two 7's in them?

Options

A. 4268
B. 6804
C. 2340
D. 1269

Answer

To solve the problem of how many 5-digit numbers contain exactly two 7's, we can break it down into two cases based on the position of the digits.

### Step-by-Step Solution:

**Step 1: Understand the constraints**
- We need to form a 5-digit number.
- The number must contain exactly two 7's.
- The first digit (leftmost) cannot be 0.

**Step 2: Case A - The first digit is 7**
- If the first digit is 7, we have the following structure: **7 _ _ _ _**.
- We need to place one more 7 in the remaining 4 positions.
- The number of ways to choose 1 position for the second 7 from the remaining 4 positions is given by \( \binom{4}{1} = 4 \).

**Step 3: Fill the remaining positions**
- After placing the two 7's, we have 3 remaining positions to fill.
- These positions can be filled with any digit from 0 to 9, except 7 (since we already have two 7's).
- Therefore, for each of the 3 positions, we have 9 choices (0-9 excluding 7).
- The total ways to fill these 3 positions is \( 9 \times 9 \times 9 = 9^3 = 729 \).

**Step 4: Calculate total for Case A**
- The total number of 5-digit numbers for Case A is:
\[ \text{Total for Case A} = \binom{4}{1} \times 9^3 = 4 \times 729 = 2916. \]

**Step 5: Case B - The first digit is not 7**
- In this case, the first digit can be any digit from 1 to 9 (it cannot be 0 or 7).
- Thus, there are 8 choices for the first digit (1-9 excluding 7).
- The structure is now: **_ _ _ _ _** (with the first digit being one of the 8 choices).

**Step 6: Place the two 7's**
- We need to place the two 7's in the remaining 4 positions.
- The number of ways to choose 2 positions for the 7's from the 4 available is given by \( \binom{4}{2} = 6 \).

**Step 7: Fill the remaining positions**
- After placing the two 7's, we have 2 remaining positions to fill.
- These positions can be filled with any digit from 0 to 9, excluding 7.
- Thus, for each of the 2 positions, we have 9 choices.
- The total ways to fill these 2 positions is \( 9 \times 9 = 9^2 = 81 \).

**Step 8: Calculate total for Case B**
- The total number of 5-digit numbers for Case B is:
\[ \text{Total for Case B} = 8 \times \binom{4}{2} \times 9^2 = 8 \times 6 \times 81 = 3888. \]

**Step 9: Combine both cases**
- The total number of 5-digit numbers containing exactly two 7's is:
\[ \text{Total} = \text{Total for Case A} + \text{Total for Case B} = 2916 + 3888 = 6804. \]

### Final Answer:
The total number of 5-digit numbers that contain exactly two 7's is **6804**.