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

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SUMMARY

The discussion focuses on calculating the number of even 3-digit numbers that can be formed using the digits 1 through 7 under two conditions: (a) each digit can only be used once and (b) digits can be reused. For the reused case, the calculation yields 147 combinations (7 options for the first digit, 7 for the second, and 3 for the last). For the non-reused case, the correct total is 90 combinations, determined by first selecting an even digit and then arranging the remaining digits accordingly.

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Dell
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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
 

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