(20 pts) how are combinations and permutations useful in finding probabilities?

So we need to write (8a^-3)^-2/3 in the simplest form. To solve the problem what we must know is that the powers can multiply or: (a^b)^c=a^(b*c). Now we can solve our problem: (8^-2/3)*(a^-3*(-2/3) ) and we can see that 8=2^3 and when we plug that in we get: 2^(3*(-2/3)) * (a^-3*(-2/3) ) and nnow we can write: (2^-2)*(a^-2)=(1/2^2)*(1/a^2)=1/(4*a^2).