If a room had to seat 20 adults and their position didnt matter is it anymore or any less than simply being 20! to work out the number of combinations?
Thanks
Hello, 200001!
I don't understand the question . . .
As written, the problem is quite silly.
If a room had to seat 20 adults and their position didn't matter,
is it simply 20! to work out the number of combinations?
I must assume quite a bit that wasn't explained.
I assume: "number of combinations" means number of seating arrangements.
I assume: there are exactly 20 chairs in the room.
I assume: there will be one person seated in any one chair.
Since their position doesn't matter, any person can sit in any chair.
Therefore, there is one seating arrangement.
It still is silly. And your example is even sillier.
The number of ways to arrange the letters $\displaystyle A~B~\&~C$ isL
a) $\displaystyle 3!=6$ in a row; b) $\displaystyle \dfrac{3!}{3}=2$ in a circle.
Your example goes along with the question “how many three letter codes can be made with the letters $\displaystyle A~B~\&~C$". That is $\displaystyle 3^3=27$.
Now what exactly is the wording of your question.
Shall we just clarify:
I am not a mathematician. Its not my area of knowledge and I have asked for some basic advice.
Im sure its great you have spent years studying but to patronise a someone who is trying to learn is simply not required.
One day you will find an aspect of life you are not at ease with and ask for help. I can imagine you would not want the same response.
I shall pass on this one thanks.