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Counting: permutations and combinations

Factorials, ordered and unordered selections.

n! counts the orderings of n things. Choosing k of them in order is n!/(n − k)!; ignoring the order divides by k! again, giving the binomial coefficient “n choose k”. These numbers are Pascal's triangle and the coefficients of (a + b)ⁿ.

Ejemplo resuelto · 5 choose 2

5 choose 2

5,\ 2

Paso a paso

  1. \binom{5}{2} = \frac{5!}{2!\,(5-2)!}

    Unordered selections: n! / (k! (n−k)!).

  2. = \frac{120}{2 \times 6} = 10

Revelar la respuesta
\binom{5}{2} = 10

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Símbolos utilizados aquí

Toque cualquier símbolo para la definición completa, una imagen, y lo que cada letra en ella significa.

Cómo: Counting: permutations and combinations

  1. Unordered selections: n! / (k! (n−k)!).

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The probability of seeing data at least this extreme if the null hypothesis were true. It is not the probability that the null hypothesis is true.

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The sample mean sits closer to the sample than the true mean does, so squared deviations from it are slightly too small on average; dividing by n − 1 instead of n corrects the bias.

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