Question 1177691: How many 3-digit even numbers can be formed from the digits 1,2,6,8, and 9 if no repetition is allowed?
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website! The resulting 3-digit number must end with one of {2,6,8} (3 choices)
If we pick the ending digit first, that leaves P(4,2) ways to arrange the other two digits (without repetition):
P(4,2)*3 = (4!/(4-2)!)*3 = 36 numbers
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Side note: if repetition is allowed, you'd have 5*5*3 = 75 numbers (qualitatively this makes sense, as you would expect more 3-digit numbers can be made if digits can be used more than once)
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