How to find the probability that Xavier wears a matching shirt and tie on Thursday The question is as follows
"When Xavier started work, he bought five shirts with matching ties, but he has since thrown the ties into
the same drawer and hung the shirts at random in his wardrobe. Each day he picks a shirt and a tie at
random to wear and then throws them in the used pile at the end of the day. Find the probability that:
He wears a matching shirt and tie on Thursday"
I do not understand how to even attempt such a question. Please help.
 A: Let $(T_1,S_1),\dots, (T_5,S_5)$ be the 5 (different) matching pairs of ties and shirts. Each day, a tie $T_i$ and a shirt $S_j$ are chosen uniformly at random among the remaining ties and shirts (the first day, $i$ is chosen uniformly in $\{1,\dots,5\}$, and so is $j$; the second day, $i$ is chosen among the four remaining numbers for the ties, etc.).
This may be hard to get a grip on, so one can think of it the following way: 2 independent permutations of $\{1,2,3,4,5\}$, denoted $(i_1,\dots,i_5)$ and $(j_1,\dots,j_5)$ are chosen independently and uniformly at random. On day $k$, the tie $T_{i_k}$ and the shirt $S_{j_k}$ are worn.
The question is: 

What is the probability that $i_4=j_4$? (i.e., that on day 4 the tie and shirt match)

A brute force way is as follows to answer this question is as follows.
$$
\mathbb{P}\{i_4=j_4\} = \sum_{n=1}^5 \mathbb{P}\{i_4=j_4=n\}
= \sum_{n=1}^5 \mathbb{P}\{i_4=n\}\mathbb{P}\{j_4=n\} = \sum_{n=1}^5 \frac{1}{5}\cdot\frac{1}{5} = \frac{1}{5}
$$
where the second equality is by independence, and the third by the fact that, for any fixed $n$, a fifth of all permutations of $\{1,\dots,5\}$ satisfy $i_4=n$.
