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.

Vinna dæmi · 5 choose 2

5 choose 2

5,\ 2

Skref fyrir skref

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

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

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

Sýna svarið
\binom{5}{2} = 10

Núna ūú Veldu vandamál, eða sláðu inn eða teikna eigin. Hvert skref, mynd, svarið falið þar til þú spyrð.

Haltu eigin vinnu þinni

Ókeypis reikningur bætir við athugasemdum við hverja kennslustund, skrá yfir það sem þú hefur lokið, leyst vandamál á einum stað og kennari sem þú getur spurt um þessa síðu.Stærðfræðin sjálf er opin öllum, skráður inn eða ekki.

Skráðu þig inn Innskráning

Tákn notuð hér

Pikkaðu á hvaða tákn sem er fyrir fulla skilgreiningu, mynd og hvað hvert bréf í því merkir.

Hvernig á að nota: Counting: permutations and combinations

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

Spurningar sem fólk spyr

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.

Meira í Statistics & Probability