Question

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

Answer 1

`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="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))`

`(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`