Jon P. answered • 09/06/15

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Let's first forget about the requirement that the number has to be odd. How many different 3-digit numbers can be formed where all the digits are different?

The first number can be any one of the 8 numbers. The second number can then be any one of the 7 that are left. That's 56 different pairs of numbers for the first two numbers.

The third number can then be any one of the 6 that are left. So there are a total of 56*6 = 336 3-digit numbers where all the digits are different.

Now, of these, how many are odd? Well, there are 8 possible digits, of which 4 are odd, so no matter what the first two digits are, half of the numbers will have an odd last digit, and half will have an even last digit. Since it's the last digit that determines whether the number is odd or even, that means that half of the 336 numbers will be odd, so the answer is 336/2 = 168.