Is there an R function to get the number of permutations of n objects take k P(n,k)?

问题

..or do I have to give

P.nk <- factorial(n) / factorial(n-k)

or

P.nk <- choose(n,k) * factorial(k)

Thank you.

回答1:

I don't know of any existing function. Your first suggestion will fail with large n. Your second idea should work fine when written as a function:

perm <- function(n,k){choose(n,k) * factorial(k)}

Then perm(500,2) will give 249500 for example.

回答2:

I think the gregmisc package provides these functions.

library(gregmisc)
permutations(n=4,r=4)

Mailing list reference: [R] permutation

回答3:

Check out nsamp(n,k,ordered=T) in the 'prob' package

回答4:

package gtools

# R version 3.5.3
install.packages("gtools")
library(gtools)

base::nrow(gtools::permutations(500,2))

result:

[1] 249500

also see combinations-and-permutations-in-r, permutation_with_replacement.R

another package prob:

base::ncol(prob::permsn(500,2))

[1] 249500

来源：https://stackoverflow.com/questions/2871763/is-there-an-r-function-to-get-the-number-of-permutations-of-n-objects-take-k-pn