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

To solve the problem, we need to count the number of 5-digit numbers $d_1d_2d_3d_4d_5$ where each digit $d_i$ is from the set {0, 1, 2, 3, 4, 5, 6, 7}, the number is greater than 50000, and $d_1 + d_5 \leq 8$.

1. Determine the possible values for $d_1$:
Since the number must be greater than 50000, $d_1$ can only be 5, 6, or 7. Thus, $d_1 \in \{5, 6, 7\}$.

2. Analyze the constraint $d_1 + d_5 \leq 8$ for each possible value of $d_1$:

• If $d_1 = 5$, then $5 + d_5 \leq 8$, which implies $d_5 \leq 3$. Therefore, $d_5 \in \{0, 1, 2, 3\}$. There are 4 possibilities for $d_5$.
• If $d_1 = 6$, then $6 + d_5 \leq 8$, which implies $d_5 \leq 2$. Therefore, $d_5 \in \{0, 1, 2\}$. There are 3 possibilities for $d_5$.
• If $d_1 = 7$, then $7 + d_5 \leq 8$, which implies $d_5 \leq 1$. Therefore, $d_5 \in \{0, 1\}$. There are 2 possibilities for $d_5$.

3. Determine the number of possibilities for $d_2$, $d_3$, and $d_4$:
Since there are no restrictions on $d_2$, $d_3$, and $d_4$ other than belonging to the set {0, 1, 2, 3, 4, 5, 6, 7}, each of them can take 8 possible values. Therefore, there are $8 \times 8 \times 8 = 8^3 = 512$ possibilities for $d_2d_3d_4$.

4. Calculate the total number of such 5-digit numbers:
We consider each case for $d_1$ separately and sum the results:

• If $d_1 = 5$, there are 4 choices for $d_5$ and 512 choices for $d_2d_3d_4$. So there are $1 \times 512 \times 4 = 2048$ such numbers.
• If $d_1 = 6$, there are 3 choices for $d_5$ and 512 choices for $d_2d_3d_4$. So there are $1 \times 512 \times 3 = 1536$ such numbers.
• If $d_1 = 7$, there are 2 choices for $d_5$ and 512 choices for $d_2d_3d_4$. So there are $1 \times 512 \times 2 = 1024$ such numbers.

Therefore, the total number of such 5-digit numbers is $2048 + 1536 + 1024 = 4608$.

Final Answer:
The total number of such 5 digit numbers is $ {4608} $.