In how many ways can we choose five people from a group of ten to form a committee?
We can tell this is a combination because the order of the people chosen doesn’t matter, just that there’s five of them chosen to form a committee.
\(_{10}C_{5}\)
\(\frac{10!}{5!\times(10-5)!}\)
\(\frac{10!}{5!\times5!}\)
\(\frac{10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{(5\times4\times3\times2\times1)\times(5\times4\times3\times2\times1)}\)
\(\frac{10\times9\times8\times7\times6}{(1)\times(5\times4\times3\times2\times1)}\)
\(\frac{30240}{1\times120}\)
\(252\)
There are two hundred and fifty two ways to choose five people from a group of ten to form a committee.
combination <- function(n, m){
result <- factorial(n)/(factorial(m)*factorial(n-m))
return(result)
}
combination(10, 5)## [1] 252