Question 1192828: 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.
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website!
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 2,4,6,8.
That's the same answer as Case 2, also 36 ways.
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Case 4: Using both 10 and 12.
Only 2 ways, 1012 and 1210.
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Total for all 4 cases: 24+36+36+2 = 98 ways.
Edwin
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
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
~~~~~~~~~~~~~~~~
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.
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.
Solved.
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