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Count flamingos


:?: Flamingos Fanny and Freddy have three offspring: Happy, Glee, and Joy. These five flamingos are to be distributed to seven different zoos so that no zoo gets both a parent and a child :(. It is not required that every zoo gets a flamingo. In how many different ways can this be done?


:!: There are two disjoint (mutually exclusive) cases we can consider that cover every possibility. We can use the sum rule to add them up since they don’t overlap!

  1. Case 1: The parents end up in the same zoo. There are 7 choices of zoo they could end up at. Then, the three offspring can go to any of the 6 other zoos, for a total of $7 · 6 · 6 · 6 = 7 · 6^3$ possibilities (by the product rule).
  2. Case 2: The parents end up in different zoos. There are 7 choices for Fanny and 6 for Freddy. Then, the three offspring can go to any of the 5 other zoos, for a total of $7 · 6 · 5^3$ possibilities.

The result, by the sum rule, is $7 · 6^3 + 7 · 6 · 5^3$ .


  • I found the above problem in “Probability & Statistics with Applications to Computing” book by Alex Tsun → Chapter 1. Combinatorial Theory → 1.1.2 Product Rule → pg-22 .
    • This book is well written, well formatted. The examples are non-trivial. All the examples come with solutions. The text is easy to understand and follow along. Highly recommended.
questions_and_answers.txt · Last modified: 2024/07/07 20:12 by raju