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
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Step-by-Step Solution
Step 1: Analyze constraints for the first digit
We need to form 5-digit numbers greater than 50000. This means the first digit (D1) 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 (D1) and the last digit (D5) should not be more than 8. We will consider each possible value for D1 and determine the allowed values for D5. The available digits for D5 are 0, 1, 2, 3, 4, 5, 6, 7.
Step 3: Calculate possibilities for each case of the first digit
For D1=5, D5 can be 0, 1, 2, 3 (since 5+D5≤8). This gives 4 choices for D5. For D1=6, D5 can be 0, 1, 2 (since 6+D5≤8). This gives 3 choices for D5. For D1=7, D5 can be 0, 1 (since 7+D5≤8). This gives 2 choices for D5. The middle three digits (D2,D3,D4) 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 8×8×8=512 choices for D2,D3,D4 for each case.
Step 4: Calculate total number of such 5-digit numbers
We sum the number of possibilities for each case. For D1=5, there are 4×512 numbers. For D1=6, there are 3×512 numbers. For D1=7, there are 2×512 numbers. We add these up to get the total count.
Step 5: Simplify the total count
Factoring out 512, we get (4+3+2)×512=9×512. Performing the multiplication, we find the total number of such 5-digit numbers is 4608.