There are 60 integers between 100 and 999 that consist of distinct odd digits.
Explanation:
To solve this problem, we need to consider the possible choices for each digit (hundreds, tens, and units) such that all digits are odd and distinct. The odd digits are 1, 3, 5, 7, and 9.
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Hundreds Digit: Since the number must be between 100 and 999, the hundreds digit can be any of the 5 odd digits (1, 3, 5, 7, or 9). So, there are 5 choices for the hundreds digit.
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Tens Digit: Once we've chosen the hundreds digit, we can't use it again for the tens digit. Since we must use odd digits, we only have 4 remaining odd digits to choose from for the tens place.
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Units Digit: Now, we've chosen two distinct odd digits for the hundreds and tens places. This leaves us with 3 remaining odd digits to choose from for the units digit.
To find the total number of integers that meet the criteria, we multiply the number of choices for each digit:
5 (choices for hundreds) 4 (choices for tens) 3 (choices for units) = 60
Therefore, there are 60 integers between 100 and 999 that consist of distinct odd digits.
