Discussion 6 - Data605 - Spring2020

2. In how many ways can we choose five people from a group of ten to form a committee?

The order we choose people here does not matter. Therefore we have a combination, as long as there are 5 people choosen to form a committee.

\({10\choose 5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!}\)

so \({10\choose 5} = \frac{10*9*8*7*6}{5*4*3*2*1} = \frac{30240}{120}\)

Therefore: \({10\choose 5} = 252\)

Let R check on our awesome work:

GiveUsCombinationOf <- function(n, m){
  got_it <- factorial(n)/(factorial(m)*factorial(n-m))
  return(got_it)
}

#Run the question!!
GiveUsCombinationOf(10, 5)

## [1] 252