The digits 1, 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.

Nice twist of similar questions.
There are 7*6*5=210 possible 3-digit numbers formed by digits 1-7 without repetition.
For numbers greater than 700, the first digit must be 7.
Of the 30, the last digit must be a 2,4, or 6 to be an even number, that leaves 3 possibilities for the last digit.
The middle digit can be any number of the remaining 5 digits.
So the number of qualifying candidates would be
1*5*3=15 (1 for the first digit, 5 choices for the second, and 3 choices for the last digit).
Probability of randomly choosing a qualifying candidate (>700 & even) is therefore 15/210=1/14

The answer is 10 bud