# How do you solve for n in P(n,4)=30[C(n-1, 3)]?

$n = 5$

#### Explanation:

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"

C_(n,k)=(n!)/((k)!(n-k)!) with $n = \text{population", k="picks}$

So let's set them equal to each other and insert what we know:

${P}_{n , 4} = 30 \left({C}_{n - 1 , 3}\right)$

(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)!)

We can now multiply the left side top and bottom by 3! to have our denominators match:

(n!)/((n-4)!)((3!)/(3!))=(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)!

We can rewrite the left side, recognizing that n! =nxx(n-1)!

3!(nxx(n-1)!)=30(n-1)!

3!n(n-1)! =30(n-1)!

divide through by (n-1)!:

3!n =30

n=30/(3!)

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