"In a tennis tournament there are 2n players. In the first round of the tournament each participant plays just once, so there are n games, each occupying a pair of players. Show that the pairing for the first round can be arranged in exactly
1x3x5x7x9x...x(2n-1) different ways."
Just to make it clear, I have access to the answer albeit very short and not as detailed as I worked on the problem. I just want to verify that my reasoning is ok (I relied heavily on geoemetry to make a sense of the product, so my answer is based on this)
We have 2n players. Take one of the players from the group and see how many combinations you can form with the other players. This means you have 2n-1 combinations with the specific player that you chose.
We have (2(n-0)-1) x ....
By what are we going to multiply 2n-1 ? To see this, let us substract 2 from 2n players because this makes sure of accounting of the fact that when we selected one of the players and associated him with the other 2n-1 players that were available, we had different pairs assigned and thus we can disregard each of these unique pairs. Thus, we actually have 2n-2 players and we repeat the same reasoning as before, we select one of these players and assign him to the different other ones, making the number of combinations with this selected player 2n-3=(2n-2)-1=2(n-1)-1.
We have (2(n-0)-1) x (2(n-1)-1) x ...
You repeat the same idea and get :
(2(n-0)-1) x (2(n-1)-1) x (2(n-2)-1) ...
We see a pattern and that all ours numbers are of the form 2n-1 and are thus odd numbers and that these odd numbers are decreasing because we are substracting 0, 1, 2, etc. from n. Thus it goes to 1.
(2(n-0)-1) x (2(n-1)-1) x (2(n-2)-1) x ... x 1
In shorthand : 2(n-a)-1 where n is random and a=0,1,2,3,...,n-1 to make sure n is bigger than a.
Like I said, this is very geometric and this my reasoning of how to solve the problem. The book solution is very very short. I can put it people want to see it.