Question 434083: Discrete Math:
What is the probability that an odd number between 1000 and 9000 contains no repeated digits?
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! We are already assuming that the number is odd, so the probability is equal to A/B, where
A is the number of odd numbers without repeated digits
B is the total number of odd numbers
We can see that B is the size of the set {1001, 1003, ..., 8999}. To count the number of elements in this set we can subtract 999 from each element to get {2, 4,..., 8000} --> {1, 2, ..., 4000} after dividing by 2. Hence, B = 4000.
To find A, we fix the units digit to be odd (5 choices), and the remaining three digits can be chosen. There are eight choices for the thousands digit (any digit except 0 or the units digit), eight choices for the hundreds digit (any digit except the thousands or units digits), and seven choices for the tens digit. This implies A = 5*8*8*7 = 2240.
Therefore the probability is A/B = 2240/4000 = 14/25.
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