C(n, k) = n! / (n-k)! k!
find n and k such that the ratios satisfy the equation
c(n, k) : c(n, k+1) : c(n, k+2) = 1: 2: 3
we sort of never learned factorials, i get the gist of how they work, but i can't even begin to comprehend what's going on here.

We have the equations:
C(n,k) / C(n,k+1) = 1/2
C(n,k+1) / C(n, k+2) = 2/3

[n! / (n-k)!k! ]/[ n! / (n- (k+1))! (k+1)! ]= 1/2

You can cancel out similar terms. The definition of factorials is this
n! = n(n -1)(n -2)(n -3)...1

So,
 (n- (k+1))! (k+1)! / (n-k)!k! = 1/2
(n- k - 1)(n - k - 1 -1)! (k+1)(k + 1 - 1)! / (n-k)!k! = 1/2
(n- k - 1)(n - k - 2)! (k+1)(k)! / (n-k)!k! = 1/2
Cancel out terms.
(n- k - 1)(n - k - 2)! (k+1)/ (n-k)!= 1/2 [eqn 1]

[n! / (n-k+1)!(k+1)! ]/[ n! / (n- (k+2))! (k+2)! ]= 2/3
(n- (k+2))! (k+2)! / (n-(k+1))!(k+1)!  = 2/3
(n- k-2)! (k+2)! / (n-k-1)!(k+1)!  = 2/3
(n- k-2)(n - k -3)! (k+2)( k+1)k! / (n-k-1)(n-k-2)(n-k-3)!(k+1)k!  = 2/3
Cancel out terms
(k+2)/ (n-k-1) = 2/3
You can solve for n in terms of k and substitute this to the fist equation which will allow you to solve for k.

2/1 or just 2. sorry if i get this wrong