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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.

Намунаи коркардашуда · 5 choose 2

Discrete-Math · Counting

5 choose 2

5,\ 2

Қадами ба қадам

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

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

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

Ҷавоби ҷавобро нишон диҳед
\binom{5}{2} = 10

Ҳоло шумо Масъаларо интихоб кунед, ё ки онро навишта ё кашед. Ҳар қадам, тасвир, ҷавоб то он даме, ки шумо пурсед, пинҳон мешавад.

Кор кардан

Дар ҳисоби ройгон ба ҳар як дарс қайдҳо илова карда мешаванд, қайди он, ки шумо чӣ кор кардаед, масъалаҳои ҳалшуда дар як ҷо ва муаллиме, ки шумо метавонед дар бораи ин саҳифа пурсед. Математика барои ҳама кушода аст, хоҳ ворид шуда бошад ё на.

Бақайдгирӣ Ворид шудан

Символы, используемые здесь

Барои муайянкунии пурраи маъно, тасвир ва маънои ҳар як ҳарф дар он, ба ҳар як аломат пахш кунед.

Чӣ тавр: Counting: permutations and combinations

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

Саволҳои маъмул

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.

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