There's a room with $25$ seats.
How many possibilities are there to place 20 different persons onto it?
I thought about calculating $${25 \choose 20} =5310 $$
But don't I have to multiply that result with $20!$ now? Wouldn't the result be too big?
And how many possibilities are there to place $10$ groups of two so that the groups are always next to each other?
Yes, you should multiply it with $20!$.
Also you can do like that: First one has $25$ choises, second one $24$, and so on till the last one who has 6 choises. So the answer is $$25\cdot 24\cdot...\cdot 7\cdot 6 = {25!\over 5!} $$ which is surprisingly the same as $${25\choose 20}\cdot 20!$$
For the second question: You have $10$ pairs and $5$ imaginary people. You can distribute them on $${15!\cdot 2^{10}\over 5!}$$ ways on $25$ chairs. We should divide it by $5!$ since any pemutation of this imaginary people is the same distibution.