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Complicated combinatorics problem

Warren and Nadia have a straight path, one meter wide and 16 meters long, from the front door to the front gate. The decide to pave it. Warren brings home 16 identical paving stones, each 1 meters square. Nadia brings home 8 identical paving stones, each 1 meter x 2 meters. Assuming that they would consider using all square, all rectangular or a combination of each, what is the probability that the combination used consists of all Warren's blocks?

Note by Joel Jablonski
4 years, 1 month ago

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I'm getting

\(\displaystyle \text{Pr(all Warren's blocks)} = \dfrac{1}{\sum_{i=0}^{8}\dfrac{(16- i)!}{i!\, (16-2\, i)!}} = \dfrac{1}{1597}\)

Gopinath No - 4 years, 1 month ago

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Nice problem. Fibonacci is the key - the number of ways of tiling a 1 by 16 path is equal to the number of ways of tiling a 1 by 15 path (by adding a 1 by 1 slab) plus the number of ways of tiling a 1 by 14 path (by adding a 1 by 2 slab), and so on. So there are 1597 (the 17th or 18th Fibonacci number, depending where you start) ways, giving a probability of 1/1597.

Now how about this question: they tile the path from the front door, and choose a slab at random each time from those they have remaining. What's the probability now that they use all of Warren's tiles? What about 8 of Warren's and 4 of Nadia's? More of Warren's than Nadia's? Warren's tiles make up more than half the length of the path?

Paddy MacMahon - 4 years, 1 month ago

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