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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.

作業例: 5 choose 2

Discrete-Math · Counting

5 choose 2

5,\ 2

ステップバイステップ

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

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

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

答えを明らかにしろ
\binom{5}{2} = 10

Symbols used here

\binom{n}{k}
binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).
n!
factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.
\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\approx
approximately equal
Equal to the precision shown, not exactly.
\sum_{k=1}^{n} a_k
summation
Add a_k for k = 1 up to n.
A \cup B,\ A \cap B,\ A \setminus B
union, intersection, difference
In either; in both; in A but not B.
\bar{x},\ \mu
sample mean, population mean
Average of the data; average of the whole population.
\sigma,\ s,\ \sigma^2
standard deviation, sample s.d., variance
Typical distance from the mean; its square.
P(A),\ P(A \mid B)
probability, conditional probability
Chance of A; chance of A given that B happened.
E[X],\ \operatorname{Var}(X)
expected value, variance
Probability-weighted average of X; its spread.
N(\mu, \sigma^2),\ z
normal distribution, z-score
The bell curve with mean μ and variance σ²; (x − μ)/σ.

How to: Counting: permutations and combinations

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

Questions people ask

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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