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.