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

The objective is to determine the count of 5-digit numbers $d_1d_2d_3d_4d_5$, where each digit $d_i$ belongs to the set {0, 1, 2, 3, 4, 5, 6, 7}. The number must exceed 50000, and the sum of the first and last digits must satisfy $d_1 + d_5 \leq 8$.

1. Identify the possible values for the first digit ($d_1$):
For the number to be greater than 50000, the first digit $d_1$ must be 5, 6, or 7. Hence, $d_1 \in \{5, 6, 7\}$.

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

• If $d_1 = 5$, the condition becomes $5 + d_5 \leq 8$, implying $d_5 \leq 3$. The possible values for $d_5$ are {0, 1, 2, 3}, totaling 4 options.
• If $d_1 = 6$, the condition becomes $6 + d_5 \leq 8$, implying $d_5 \leq 2$. The possible values for $d_5$ are {0, 1, 2}, totaling 3 options.
• If $d_1 = 7$, the condition becomes $7 + d_5 \leq 8$, implying $d_5 \leq 1$. The possible values for $d_5$ are {0, 1}, totaling 2 options.

3. Determine the number of possibilities for the middle digits ($d_2, d_3, d_4$):
As there are no specific restrictions on $d_2$, $d_3$, and $d_4$ beyond belonging to the set {0, 1, 2, 3, 4, 5, 6, 7}, each of these digits has 8 potential values. Consequently, there are $8 \times 8 \times 8 = 8^3 = 512$ combinations for the sequence $d_2d_3d_4$.

4. Calculate the total count of qualifying 5-digit numbers:
The total count is obtained by summing the possibilities for each case of $d_1$:

• For $d_1 = 5$: 1 choice for $d_1$, 4 choices for $d_5$, and 512 choices for $d_2d_3d_4$. This gives $1 \times 512 \times 4 = 2048$ numbers.
• For $d_1 = 6$: 1 choice for $d_1$, 3 choices for $d_5$, and 512 choices for $d_2d_3d_4$. This gives $1 \times 512 \times 3 = 1536$ numbers.
• For $d_1 = 7$: 1 choice for $d_1$, 2 choices for $d_5$, and 512 choices for $d_2d_3d_4$. This gives $1 \times 512 \times 2 = 1024$ numbers.

The aggregate total of such 5-digit numbers is $2048 + 1536 + 1024 = 4608$.

Final Answer:
The total count of the specified 5-digit numbers is $ {4608} $.