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Answer :
Final Answer:
There are 1796 numbers from 10000 to 99999 that have two repeated digits and the other three digits the same but of a different value.
Explanation:
To find the numbers that meet the given criteria, we can consider the following cases:
1. The repeating digits are in the tens of thousands and thousands places (e.g., 22345, 55999).
2. The repeating digits are in the thousands and hundreds places (e.g., 32234, 99599).
3. The repeating digits are in the hundreds and tens places (e.g., 43223, 79995).
4. The repeating digits are in the tens and ones places (e.g., 54222, 89999).
For each of these cases, we can determine the number of possibilities:
- For case 1, there are 9 choices for the repeated tens digit (1 to 9) and 10 choices for the repeated thousands digit (0 to 9, excluding the tens digit itself). This gives us 9 * 10 = 90 possibilities.
- For cases 2 and 3, there are 9 choices for the repeated hundreds digit, 9 choices for the repeated thousands digit, and 10 choices for the remaining digit. This gives us 9 * 9 * 10 = 810 possibilities for each case.
- For case 4, there are 9 choices for the repeated tens digit and 10 choices for the repeated ones digit. This gives us 9 * 10 = 90 possibilities.
Adding up the possibilities from all four cases, we get a total of 90 + 810 + 810 + 90 = 1800 possibilities. However, we need to subtract the cases where all digits are the same (e.g., 11111), which gives us 4 possibilities.
Therefore, the total number of numbers that meet the criteria is 1800 - 4 = 1796.
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