Basic Probability - compound experiment 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.
 A: You need to figure out all the ways 6 sets of clothing can be distributed among six people so that person 1 gets a specific set of clothing (suit + tie). You then divide by all possible combinations to get a probability.
There are two ways you could think about this:


*

*Thinking only of person 1, how many ways can he get exactly one specific set of clothing? (Put another way: how many ways can you distribute one thing to one person?) Then multiply by the number of ways the remaining five sets of clothing can be distributed among the remaining five people. Then divide by the total number of ways all six sets of clothing can be distributed among all six people.

*What's the probability of a specific person getting a specific set of clothing, if there are N sets of clothes and N people? 
Both approaches give the same answer. (Which is to say that you were right when you asked "How can these two problems have different solutions in the first place?") If you look carefully at the answer choices, you'll see that two of them  work out to be the same :-)
