Question

The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is:

• A. 5719
• B. 4608
• C. 5720
• D. 4607

Hint:

Utilize the properties of digits and place values effectively in counting problems. Understanding the constraints on digits at specific positions can greatly simplify the calculation.

Answer 1

The Correct Option is B

Step 1: Determine the first digit restrictions.

The first digit must be \( \geq 5 \) to ensure the number is greater than 50000. This restricts the first digit to 5, 6, or 7.

Step 2: Calculate possible combinations for each first digit.

For each valid first digit \( d_1 \) (5, 6, or 7), determine possible last digits \( d_5 \) such that their sum \( d_1 + d_5 \leq 8 \):

\[ \begin{aligned} \text{For } d_1 = 5: & \quad \text{Possible } d_5 \text{ are } 0, 1, 2, 3 \quad \text{(4 choices)} \\ \text{For } d_1 = 6: & \quad \text{Possible } d_5 \text{ are } 0, 1, 2 \quad \text{(3 choices)} \\ \text{For } d_1 = 7: & \quad \text{Possible } d_5 \text{ are } 0, 1 \quad \text{(2 choices)} \end{aligned} \]

Step 3: Count total combinations.

Each of the middle three digits (\(d_2, d_3, d_4\)) can be any of the 8 digits (0-7). Calculating the combinations for each case:

\[ \begin{aligned} \text{For } d_1 = 5: & \quad 4 \times 8^3 = 2048 \\ \text{For } d_1 = 6: & \quad 3 \times 8^3 = 1536 \\ \text{For } d_1 = 7: & \quad 2 \times 8^3 = 1024 \end{aligned} \]

Step 4: Sum over all valid first digits.

\[ \text{Total combinations} = 2048 + 1536 + 1024 = 4608. \]

Conclusion:

The total number of such 5-digit numbers greater than 50000, formed under the given constraints, is 4608.