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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.
Eżempju maħdum · 5 choose 2
Pass b'pass
- \binom{5}{2} = \frac{5!}{2!\,(5-2)!}
Unordered selections: n! / (k! (n−k)!).
- = \frac{120}{2 \times 6} = 10
Jiżvelaw it-tweġiba
Issa inti Agħżel problema, jew ittajpja jew iġbed tiegħek stess.Kull pass, stampa, it-tweġiba moħbija sakemm inti tistaqsi.
Kont b'xejn iżid noti fuq kull lezzjoni, rekord ta' dak li lest, problemi solvuti tiegħek f'post wieħed, u għalliem inti tista' tistaqsi dwar din il-paġna.Il-matematika nnifisha hija miftuħa għal kulħadd, log fil jew le.
Irreġistra issa LoginSimboli użati hawnhekk
Taptap kwalunkwe simbolu għad-definizzjoni sħiħa, stampa, u dak li kull ittra fiha tfisser.
Kif għandek tuża: Counting: permutations and combinations
- Unordered selections: n! / (k! (n−k)!).
Mistoqsijiet li n-nies jistaqsu
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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