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

Step 1: Analyze constraints for the first digit

We need to form 5-digit numbers greater than 50000. This means the first digit ($D_1$) cannot be 0, 1, 2, 3, or 4. Since the available digits are 0, 1, 2, 3, 4, 5, 6, 7, the first digit must be 5, 6, or 7.

Step 2: Analyze constraints for the last digit based on the first digit

The problem states that the sum of the first digit ($D_1$) and the last digit ($D_5$) should not be more than 8. We will consider each possible value for $D_1$ and determine the allowed values for $D_5$. The available digits for $D_5$ are 0, 1, 2, 3, 4, 5, 6, 7.

Step 3: Calculate possibilities for each case of the first digit

For $D_1 = 5$, $D_5$ can be 0, 1, 2, 3 (since $5+D_5 \le 8$). This gives 4 choices for $D_5$. For $D_1 = 6$, $D_5$ can be 0, 1, 2 (since $6+D_5 \le 8$). This gives 3 choices for $D_5$. For $D_1 = 7$, $D_5$ can be 0, 1 (since $7+D_5 \le 8$). This gives 2 choices for $D_5$. The middle three digits ($D_2, D_3, D_4$) can be any of the 8 available digits (0-7), as repetition is allowed and there are no further restrictions on them. So, there are $8 \times 8 \times 8 = 512$ choices for $D_2, D_3, D_4$ for each case.

Step 4: Calculate total number of such 5-digit numbers

We sum the number of possibilities for each case. For $D_1=5$, there are $4 \times 512$ numbers. For $D_1=6$, there are $3 \times 512$ numbers. For $D_1=7$, there are $2 \times 512$ numbers. We add these up to get the total count.

Step 5: Simplify the total count

Factoring out 512, we get $(4+3+2) \times 512 = 9 \times 512$. Performing the multiplication, we find the total number of such 5-digit numbers is 4608.

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