In how many ways can a team of 20 hockey players be accommodated in 10 two-person hotel rooms? Assume that the order of assigning the rooms does not matter.
+202 First room: \(\binom{20}{2}\)
Second room: \(\binom{18}{2}\)
And so on. Use multiplication I think. Someone check my work because knowing these types of problems there's usually a catch or my method is completely wrong.
+202 The so on is meant to say that this continues until you get down to the last room.
+118687 Thanks Impasta,
I am not totally sure either but Impasta's answer looks logical.
I think Impasta's method assumes that all the hotel rooms are all different in some way.
If they are identical and the only thing of importance is the way the people are paired off and the number would be much lower.
(I think)
Maybe then you might divide by 10! But I am really unsure.
