Bifalgorithm post · Brisk · Chapter 1

How many ordered outcomes are possible for first, second, and third place?

How many ordered outcomes are possible for first, second, and third place? (Brisk · Chapter 1) — "To encourage rivalry and gluttony, we’re going to give a spec…

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Contar los posibles resultados de un concurso de 1.º, 2.º y 3.º lugar
Los certificados buscan fomentar la rivalidad y la glotonería.
El concurso asigna certificados a quienes ocupen los tres primeros lugares.
El problema es una permutación parcial porque importa el orden de los ganadores.
Hay 11 posibles ganadores del primer lugar.
Para cada ganador, hay 10 posibles ocupantes del segundo lugar.
Para cada pareja de primeros lugares, hay 9 posibles ocupantes del tercer lugar.
Las posibilidades se multiplican: 11 × 10 × 9.
El resultado es 990; el texto también lo presenta como «11^3».
El resultado parece sorprendentemente alto.

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How many ordered outcomes are possible for first, second, and third place?

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"To encourage rivalry and gluttony, we’re going to give a special certificate to the child who collects the most candy at the end of the night. And while we’re at it, we’ll give 2nd-place and 3rd-place certificates as well. How many different ways could our 1st-2nd-3rd contest turn out? This is a partial permutation: there are eleven possible winners, and ten possible runners-up for each possible winner, and nine possible 3rd-placers for each of those top-twos. The answer is therefore 11^3, or 990. Wow! I wouldn’t have guessed that high."

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