maths.free › Statistics & Probability › Counting: permutations and combinations
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.
مثال عملي · 5 choose 2
خطوة خطوة
- \binom{5}{2} = \frac{5!}{2!\,(5-2)!}
Unordered selections: n! / (k! (n−k)!).
- = \frac{120}{2 \times 6} = 10
كشفت الإجابة
الآن أنت اختار مشكلة، أو طبعها أو رسمها بنفسك، كل خطوة، صورة، الإجابة مخفية حتى تسأل.
الرموز المستخدمة هنا
لمس أي رمز للحصول على التعريف الكامل، والصورة، وما يعنيه كل حرف في ذلك.
كيف: Counting: permutations and combinations
- Unordered selections: n! / (k! (n−k)!).
أسئلة يطرحها الناس
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.
أكثر في Statistics & Probability
Sampling and dataDescribing data with graphsMean, median and modeProbabilityDiscrete random variablesContinuous random variablesThe normal distributionThe central limit theoremConfidence intervalsHypothesis testingComparing two samplesChi-square testsLinear regression and correlationANOVA and the F distribution