Question 1192828
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How many 4-digit numbers can be formed by using 2, 4, 6, 8, 10,12 without repetition of digits? 
if it is a combination
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            If the problem is about writing numbers base  10  (as I understand it from the context),

            then the analysis by  Edwin is not precisely accurate,

            so  I  came to bring a correct solution.



<pre>
Case 1: Using only 2,4,6,8

There are P(4,4) = (4)(3)(2)(1) = 4! = 24 ways to arrange them.

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Case 2: Using 10, and 2 others digits from 2,4,6,8

There are 3 basic ways for case 2: 
10 _ _,   _ 10 _, and _ _ 10

For each of those 3, there are P(4,2) = (4)(3) = 12 ways to arrange
two others in the 2 blanks.

So that's (3)(12) = 36 ways.

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Case 3: Using 12, and 2 others digits from 4,6,8  

    (notice that using dgit 2 is prohibited in this case)


There are 3*(3*2) = 3*6 = 18 ways in this case.

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Case 4:  Using both 10 and 12 simulataneously is PROHIBITED.

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Total for all 4 cases: 24 + 36 + 18 + 0 = 78 ways.
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Solved.