For $\displaystyle 1 \le k \le 24$ you can definitely show that the master must have played exactly $k$ games on some set of consecutive days, using pigeonhole principle.
Suppose the total number of games the master has played till the end of day $\displaystyle j$ is $\displaystyle g_j$.
Now consider the $\displaystyle 154$ numbers: $\displaystyle g_1, g_2, \dots, g_{77}, g_1 + k, g_2 + k, \dots, g_{77} + k$
These are a set of $\displaystyle 154$ numbers between $\displaystyle 1$ and $\displaystyle 132+k$.
For $\displaystyle k \lt 22$, two of these must be equal. Since $\displaystyle g_i \neq g_j$ (at least one game a day) we must have that $\displaystyle g_j + k = g_i$ for some $\displaystyle i,j$.
For $\displaystyle k=22$ we must have that the numbers are $\displaystyle 1,2, \dots, 154$, in which case, the first (and last) $\displaystyle 22$ days, the master must have played $\displaystyle 1$ game everyday.
For $\displaystyle k=23$, we can assume $\displaystyle g_i \neq 23$, and since $g_i \ge 1$, we have $g_i + k \neq 23$.
Thus by an argument similar to above, we must have have $\displaystyle 154$ numbers taking all values in $\displaystyle 1, 2, \dots, 155$, except $\displaystyle 23$ and the master must have played $\displaystyle 1$ game each of the last $\displaystyle 23$ days.
For $\displaystyle k=24$, you can show that the master must have played $\displaystyle 1$ game the first $\displaystyle 23$ days (after eliminating one of the numbers in $\displaystyle 133, 134 \dots$), then a big number of games the next, violating the $\displaystyle 12$ games per week restriction (this is where we actually used that restriction for a specific week).
We might be able to use this kind of argument to show for $\displaystyle k$ close to $\displaystyle 24$, but for larger $\displaystyle k$, I am guessing that you can find a set of games which will miss that (perhaps a computer search will help there).
It is not really a question of "versus". They are often applied together.
In the first lot of problems, you are counting ways to select elements from sets (collections of distinct elements). Sometimes you are also counting ways to arrange them. That is combinations and permutations (respectively).
In the second lot of problems, you are performing selections from multiple sets, in sequence. Thus each task can be divided into a series of independent sub-tasks; hence the Universal Principle of Counting is also used.
Best Answer
There are $18$ courses that are silver or bronze, with twice as many bronze as silver. So there are $2$ gold, $6$ silver, and $12$ bronze.
For the first question, there are $8$ choices.
For the second question, the place for the first round can be chosen in $12$ ways. For each of these choices, the place for the second round can be chosen in $6$ ways. And for each way of choosing where to play the first two rounds, there are $2$ choices for the location of the third round.
Thus the total number of choices is $(12)(6)(2)$.
Remark: We did not worry about whether "permutation" or "combination" was involved. We just solved the problem.
The only thing that was used is what is sometimes called the Multiplication Principle. If one can do Task $1$ in $a$ ways, and for each of thse ways we can do Task $2$ in $b$ ways, then we can do Tasks $1$ and $2$ in $ab$ ways. One should not memorize this as a rule, it should just be obvious. If Alicia has $a$ children, and each child has $b$ children, then Alicia ends up with $ab$ grandchildren. And if each grandchild has $c$ children, Alicia ends up with $abc$ great-grandchildren. And so on.
I will take the risk of confusing you, and say that the problem has elements of both permutations and combinations!
We want to count the number of sequences $(A,B,C)$, where $A$ is a bronze golf course, $B$ is a silver, and $C$ is a gold. Sequences sound like permutations.
But to put it another way, we want to choose a collection of three golf courses, with some restrictions: one of the choices has to be gold, another silver, another bronze. "Choose" sounds like combinations.