You are correct that the number of possible numbers is the number of permutations, which is 6!/3! = 120
Of those numbers, 1/3 will start with 7 or 9. Those numbers will exceed 700 and the others will not. So the answer is 40.
Assume that 3 digits are selected at random from the set {1,3,5,6,7,9} and are arranged in random order.
What is the probability that the resulting 3-digit number is less than 700?
I know the sample space is P(6,3) but that is all I know. Help!
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