How Do You Calculate Permutations for Complex Problems?

  • Thread starter Thread starter chwala
  • Start date Start date
  • Tags Tags
    Permutations
AI Thread Summary
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
Gold Member
Messages
2,827
Reaction score
415
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:
 
Physics news on Phys.org
You can easily write down all fourteen numbers that end in ##2## and check your counting for each category.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top