Solution yields 120 possibilities

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SUMMARY

The problem discussed involves calculating the number of distinct arrangements of the letters in the word "BANANAS" such that no two 'A's are adjacent. The solution presented yields 120 valid permutations. Participants confirm this result by suggesting an alternative approach that involves first counting the permutations where at least two 'A's are together and then subtracting from the total permutations of the letters.

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  • Familiarity with permutations and combinations
  • Knowledge of the principle of inclusion-exclusion
  • Basic grasp of factorial notation
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  • Study the principle of inclusion-exclusion in combinatorics
  • Learn about permutations of multiset arrangements
  • Explore advanced counting techniques in combinatorial mathematics
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Mathematicians, students studying combinatorics, educators teaching counting principles, and anyone interested in solving permutation problems with constraints.

hypermonkey2
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what would a solution to the following problem look like?
How many "words" (distinct orderings of letters) can you make of the word BANANAS in which no As are beside each other. This may turn out to be a simple counting problem, and i apologize if i waste anyones time. My solution yields 120 possibilities i think. am i correct? thanks.
 
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I also get 120. It's easy if you first count the number of permutations of BANANAS with at least 2 As beside each other, etc.
 

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