To solve this, we need to first know the general formulas for permutations and combinations:
P_(n,k)=(n!)/((n-k)!); n="population", k="picks"Pn,k=n!(n−k)!;n=population,k=picks
C_(n,k)=(n!)/((k)!(n-k)!)Cn,k=n!(k)!(n−k)! with n="population", k="picks"n=population,k=picks
So let's set them equal to each other and insert what we know:
P_(n,4)=30(C_(n-1,3))Pn,4=30(Cn−1,3)
(n!)/((n-4)!)=(30(n-1)!)/((3)!((n-1)-3)!)n!(n−4)!=30(n−1)!(3)!((n−1)−3)!
We can combine terms in the right hand side denominator:
(n!)/((n-4)!)=(30(n-1)!)/((3)!(n-4)!)n!(n−4)!=30(n−1)!(3)!(n−4)!
We can now multiply the left side top and bottom by 3!3! to have our denominators match:
(n!)/((n-4)!)((3!)/(3!))=(30(n-1)!)/((3)!(n-4)!)n!(n−4)!(3!3!)=30(n−1)!(3)!(n−4)!
(3!n!)/(3!(n-4)!)=(30(n-1)!)/((3)!(n-4)!)3!n!3!(n−4)!=30(n−1)!(3)!(n−4)!
With the denominators the same, we can multiply through by it (and thus eliminate them) and thereby equate the numerators:
3!n! =30(n-1)!3!n!=30(n−1)!
We can rewrite the left side, recognizing that n! =nxx(n-1)!n!=n×(n−1)!
3!(nxx(n-1)!)=30(n-1)!3!(n×(n−1)!)=30(n−1)!
3!n(n-1)! =30(n-1)!3!n(n−1)!=30(n−1)!
divide through by (n-1)!(n−1)!:
3!n =303!n=30
n=30/(3!)n=303!
color(blue)(ul(bar(abs(color(black)(n=30/6=5))))
Let's check this:
(5!)/((5-4)!)=(30(5-1)!)/((3)!((5-1)-3)!)
(5!)/(1!)=(30(4!))/((3)!((4)-3)!)
5! =(30(4!))/((3)!(1!)
5! =(30(4!))/6
5! =5(4!)
5! =5! color(white)(000)color(green)sqrt
