Question: A large pool of adults earning their first driver’s license includes 50% low-risk drivers, 30% moderate-risk drivers, and 20% high-risk drivers. Because these drivers have no prior driving record, an insurance company considers each driver to be randomly selected from the pool.

This month, the insurance company writes 4 new policies for adults earning their first driver’s license.

What is the probability that these 4 will contain at least two more high-risk drivers than
low-risk drivers?

This is the solution included:
Let
X = number of low-risk drivers insured
Y = number of moderate-risk drivers insured
Z = number of high-risk drivers insured
f(x, y, z) = probability function of X, Y, and Z
Then f is a trinomial probability function, so
Pr[z≥x+2]=f(0,0,4)+f(1,0,3)+f(0,1,3)+f(0,2,2)
=(.2)^4+4(.5)(.2)^3+4(.3)(.2)^3+(4!/2!2!)(.3)^2(.2)^2
=.0488

I get where all the decimals are coming from, but why are they multiplied by 4 and then (4!/2!2!) ?