Consider three positive integers x, y, and z that satisfy this equation:
comb(x,y)*comb(y,z) = 2*comb(x,z)
Does the above equation possess a finite number of solutions?
If so, determine all possible solutions of the above mentioned equation.
If not - then derive, with proof, the possible relationship between x, y, and z.
Note: comb(m,n) is the number of ways of choosing n unordered outcomes from m possibilities and defined as:
m!
comb(m,n) = ---------
n!(m-n)!
It is also known as binomial coefficient and read as "m choose n".
We have C(x,y) * C(y,z) = 2 * C(x,z). so lets start by substituting the definition:
x!/(y!(x-y)!) * y!/(z!(y-z)!) = 2 * x!/(z!(x-z)!)
Now move everything except the constant 2 to the left side and simplify:
(x-z)!/((x-y)!(y-z)!) = 2
The left side is just C(x-z, y-z), or equivalently C(x-z, x-y).
The only way for C(a,b) to equal 2 is when a=2 and b=1. Then x-z=2 and y-z=1.
So then x, y, z are three consecutive integers in descending order.