Question 1143267
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There are 7 digits from which 3 must be selected, and order matters:
(a) P(7, 3) = 7!/(7-3)! = 7!/4!  = 7*6*5 = 210

You are only supposed to post one problem per-post, but I'm in a good mood so I will give hint for (b):

Hint for (b):  how many numbers end with 2,4,6, or 8?   Figure out how many there are, then the probability is that number divided by the answer from (a).

If you look at the last digit of the 3 digit number, there are C(4,1) ways to select an even digit.  Once the ending digiit is known to be even, the other two digits can be even or odd.   So C(4,1) must be multiplied by P(6,2) which is the number of ways of arranging the other 2 digits out of the 6 remaining digits.

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For part (b), the value 1/6 is incorrect.  The correct answer is 4/7 (=120/210). Also notice that there are 4 even and 3 odd digits, so 4/7 are even.  Say you pick the ending digit first, that has a 4/7 chance of being even.  So simple!