How many ways could the guests arrange themselves on a four person couch

If I understand the question correctly; we have one couch for four people and 33 guests so 29 guests have to stand.

On the first place on the couch 33 people can sit, on the second place only 32 can sit (because one is already sitting on the first place), and so on...

So $33\times 32\times 31\times 30=982080$ ways.

From the 33 guests, choose 4 to sit on the four-person couch: ${33 \choose 4}$. After that, arrange those 4 guests on the couch in all possible ways: $4!$. Hence the final answer is: $$ {33 \choose 4} 4! $$

This is a case of permutations without repetition. To visualize this better, imagine you have a total of $n$ balls and you want to place them in $r$ boxes. Each box can contain only $1$ ball. Clearly, the number of boxes filled cannot surpass $n$, since $n \geq r$.

You want to see the number of different ways you could do that. In other words, you want to find the number of permutations. For the $1^{st}$ box, you could place any one of the balls. Hence, there are $n$ options. For the $2^{nd}$ box, you have $1$ fewer ball, so there will be $(n-1)$ options. For the $3^{rd}$ box, there will be $(n-2)$ options. Keep repeating until you reach the $r^{th}$ box, where there will be $(n-(r-1))$ options. Let’s say $P$ represents the total number of permutations for $n$ choose $r$. $$P_{(n, r)} = n\cdot(n-1)\cdot(n-2)\cdot(n-3)...\cdot(n-(r-1)) = \frac{n!}{(n-r)!}$$ In your question, there are $33$ guests and a $4$-person couch, so $n = 33$ and $r = 4$. $$P_{(33, 4)} = \frac{33!}{(33-4)!} = \frac{33!}{29!}$$ All factors from $29$ to $1$ will cancel. $$P_{(33, 4)} = 33\cdot 32\cdot 31\cdot 30$$ $$\boxed{P_{(33, 4)} = 982080}$$