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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.

Isibonelo esisebenza · 5 choose 2

5 choose 2

5,\ 2

Isigaba

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

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

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

Bonisa impendulo
\binom{5}{2} = 10

Manje Khetha izinkinga, noma ubhale noma udwebe zakho. Isigaba ngasinye, isithombe, impendulo ifihlakele kuze kube ubuza.

Qhubeka usebenza

I-akhawunti emahhala ifaka iziphawulo kulesifundo ngasinye, irekhodi lokuthi yini oqedile, izinkinga zakho ezixazululiwe endaweni eyodwa, nodokotela owaziyo ongabuza mayelana nalekhasi. I-matematics ivulekile kubo bonke, noma ngabe babhalise noma cha.

Bhala Ngena

Amaphawu asetshenziswa lapha

Chofoza noma yiluphi uphawu lwesibonakaliso ukuze uthole ukuchaza okuphelele, isithombe, kanye nokuthi iyiphi inhlamvu ekhona ekhuluma ngaso.

Indlela: Counting: permutations and combinations

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

Imibuzo abantu bebuza

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.

Okuningi Statistics & Probability