Question 1058256: The digits 2, 3, 4, 5, 6, and 7 are randomly arranged to form a three-digit number. (Digits are not repeated.) Find the probability that the number is even and greater than 700.
Answer by solve_for_x(190)   (Show Source):
You can put this solution on YOUR website! Since the number must be greater than 700, there is only one choice for the first digit: 7
Since the number must be even, there are only 3 choices for the last digit: 2, 4, 6
And since there are no repeated digits, there are only 4 choices for the middle digit (cannot be 7, or the same as the last digit).
The total number of possible arrangements is then 1 * 4 * 3 = 12
The possibilities are:
724
726
732
734
736
742
746
752
754
756
762
764
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