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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.

Ýüklenen mysal · 5 choose 2

5 choose 2

5,\ 2

Adım adım

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

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

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

Jawaby görkez
\binom{5}{2} = 10

Indi sen Bir meseläni saýla, ýa-da ýaz ýa-da özüňi surata çek. Her ädim, bir resim, jogap sorag berlençä gizlenýär.

Özüň işle

Mugt hasap her bir ders üçin notlar goşýar, tamamlanan zadyňyzyň ýazgysyny, çözen problemaňyzy bir ýerde, we bu sahypa barada sorap biljek mugallymy. Matematika özüçe her kim üçin açyk, girseň ýa girmeseň.

_Gün _Giriş

Şu ýerde ulanylan simwollar

Eň täze düşündirme, surat, we her harpyň manysy üçin her bir simwoly basyň.

Nädip: Counting: permutations and combinations

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

Adamlar soraýan soraglar

Mean or median, which should I use?

Median when the data have outliers or a long tail (incomes, house prices); mean when the data are roughly symmetric and you want every value to count. Report both if they disagree. The gap is itself information.

What does a p-value actually say?

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.

Why divide by n − 1 for the sample variance?

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.

_Ýaşa Statistics & Probability