Data Management: Premutations Problem With identical Items

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SUMMARY

The problem involves calculating the number of ways to fill four positions with ten students, including two indistinguishable Norman twins. The correct approach involves using permutations with identical items, leading to the formula n!/(a!b!...) for identical items. The total number of ways to fill the positions is calculated as 4032, which differs from the initial calculation of 5040 due to the indistinguishability of the twins. The breakdown into cases for the twins' involvement is essential for accurate computation.

PREREQUISITES
  • Understanding of permutations and combinations
  • Familiarity with factorial notation (n!)
  • Knowledge of the concept of indistinguishable items in combinatorics
  • Ability to apply the formula nPr = n!/(n-r)! for permutations
NEXT STEPS
  • Study the concept of permutations with identical items in combinatorial mathematics
  • Learn how to apply the factorial function in various combinatorial problems
  • Explore case analysis techniques for solving complex permutation problems
  • Review examples of indistinguishable objects in combinatorial contexts
USEFUL FOR

Students studying combinatorics, educators teaching permutation concepts, and anyone interested in solving complex counting problems involving indistinguishable items.

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Homework Statement


Ten students have been nominated for positions of secretary, treasurer, social convenor, and fundraising chair. In how many ways can these positions be filled if the Norman twins are running and plan to switch positions on occasion for fun since no one can tell them apart?

txt book answer:4032

Homework Equations


n! factorials

nPr= n!/(n-r)! premutations

n!/(a!b!c!...) premutations with identical items


The Attempt at a Solution


10P4=5040 totals ways nominations can be picked

i don't know how to find the number of ways twins can be picked/ answer
 
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I guess we are to treat the twins as indistinguishable. So try breaking the problem up into three cases:
(1) No twin holds a position;
(2) Exactly one twin holds a position;
(3) Both twins hold positions.

I think I should warn you that I don't get the book answer. So I'm going to stick my neck out and say that the book is wrong. Of course, it may be that I have misinterpreted the problem. I do that a lot.
 

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