How Do You Calculate Permutations for Complex Problems?

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The discussion focuses on calculating permutations for complex problems, specifically arriving at a total of 28 ways to fill in blanks in a sequence. Two different approaches are presented, both leading to the same result. The first method calculates permutations based on specific combinations, while the second method uses a formula involving permutations and combinations to arrive at the same total. The importance of verifying the counting for each category is emphasized, suggesting a thorough check of all possibilities. Overall, the calculation of permutations in this context is confirmed to yield a consistent result of 28.
chwala
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Homework Statement
How many even numbers less than ##500## can be formed using the digits; ##[1,2,3,4,5]?## Each digit may be used only once in any number
Relevant Equations
Permutations
Tricky questions ;

Ok in my approach;

[..., .... , ...2...] This can be filled in ##3×3×1=9## ways

[..., .... , ...4...] This can be filled in ##3×3×1=9## ways

[.... , ...4...] This can be filled in ##4×1=4## ways

[.... , ...2...] This can be filled in ##4×1=4## ways

[2] This can be filled in ##1## way

[4] This can be filled in ##1## way

Therefore number of ways is ##9+9+4+4+1+1=28## ways...

Alternatively, i was thinking of the following possibilities;

4,..., 2 The blank can be filled in ##3P1×1## ways

1,..., 2/4 The blank can be filled in ##3P1×2## ways

2,..., 4 The blank can be filled in ##3P1×1## ways

3,..., 2/4 The blank can be filled in ##3P1×2## ways

This will give us; ##3P1×6=18## ways +

[.... , ...4...] This can be filled in ##4×1=4## ways

[.... , ...2...] This can be filled in ##4×1=4## ways

[2] This can be filled in ##1## way

[4] This can be filled in ##1## way

##18+10=28##

The text gives solution only as ##28.##

Your input highly appreciated. This is one of my less favorite topic...:wink:
 
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You can easily write down all fourteen numbers that end in ##2## and check your counting for each category.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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