How many distinct 3 digit odd integers are there? Distinct meaning none of the digits repeat.
first digit: 9 choices (1-9)
then, 2nd digit: 9 choices (now 0 is allowed)
then, 3rd digit: 8 choices
so, 9*9*8
It has to be odd, so the 3rd digit has to be 1, 3, 5, 7, or 9. So there are only 5 options. Moreover, you do not consider the cases in which the digits do not repeat.
you got me on the odd value, but I do account for the non-repeatability of digits.
If they can be repeated, then there would be 9*10*10 possibilities for 3 digits.
My counting forces the preceding choices not to be used again.
Since the number must be odd, things get a bit more complicated, because we don't know whether the previous choices were odd or even.
You seem to have a handle on the problem. What is your reasoning?
To find the number of distinct 3-digit odd integers with non-repeating digits, we need to consider the possible choices for each digit.
First Digit: Since this is a 3-digit number, the first digit cannot be zero. So we have 9 choices: {1, 2, 3, 4, 5, 6, 7, 8, 9}.
Second Digit: Once we have chosen a digit for the first position, we have 9 remaining digits to choose from (since none of the digits can repeat). However, since the number has to be odd, the second digit cannot be even. So we have 5 choices for the second digit: {1, 3, 5, 7, 9}.
Third Digit: After choosing the first two digits, we have 8 remaining digits to choose from. However, the third digit cannot be even and cannot be the same as the first or second digit chosen. So we have 7 choices for the third digit.
To find the total number of distinct 3-digit odd integers, we multiply the number of choices for each digit: 9 choices for the first digit × 5 choices for the second digit × 7 choices for the third digit = 315.
Therefore, there are 315 distinct 3-digit odd integers with non-repeating digits.