Combinatorics: Even 3-Digit Numbers from 1-7 with Restrictions

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given the numbers 1 2 3 4 5 6 7, how many even 3 digit numbers can be made from these 7 digits
a) if each number can only be used once
b) if each number can be rused


for b) what i did was:
for the first digit i have 7 options, for the second still 7, since i can reuse, and for the 3rd only 3 since the last digit needs to be even

7*7*3=147

but for a) i am stuck, for the first i have 7 options, for the second i have 6 since i have used one already, but how many do i have for the final digit, i would think 3, but what if the 1st 2 were also even, then i only have 1 even left? the correct answer is 90 but i cannnot get it
 
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try counting from the even digit first... ie pick the smallest digit first then so on...
 
nice, thanks

3*5*6
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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