I am trying to understand this basic probability problem from Analytics Vidhya AV Casino lesson:
Each person can choose whether to wear a black tux or a grey tux, and whether to wear blue, yellow or green coloured tie.
What is the probability that person 1 gets a grey blue set if all possible sets are distributed to 6 different people such that no set repeats?
$\mathrm{A.}\quad\frac{1}{_{12}\mathrm{ P }_6}\\\mathrm{B.}\quad \frac{1}{6!}\\\mathrm{C.}\quad\frac{5!}{6!}\\\mathrm{D.}\quad\frac{5!}{_{12}\mathrm{P}_6}$
What is the probability that person 1 gets a black blue set if all possible sets are distributed to 6 different people such that no set repeats?
$\mathrm{A.}\quad\frac{1}{6}\\\mathrm{B.}\quad\frac{5!}{(6!-1)}\\\mathrm{C.}\quad\frac{1}{5!}\\\mathrm{D.}\quad\frac{1}{(6!-3)}$
My initial thoughts were that there must be wrong with my assumptions. How can these two problems have different solutions in the first place? The probability of any one set occuring is equal to the probability of any other. Furthermore my answers were incorrect on both questions. I tried and tried to understand the reasoning but could not get it. Any help is appreciated.