Dividing a set of n elements in r groups where groups are of size n1,n2,n3... and n1+n2+n3+ ...(uo to r terms ) = n

Question

(Reference- A first course in probablity by Sheldon Ross [Example 5b and 5c of chapter 1])

Q1. Ten children are to be divided into an A team and a B team of 5 each. The A team will play in one league and the B team in another. How many different divisions are possible?
Answer:10!/(5! * 5!)

Q2.In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible?
Answer:(10!/(5! * 5!))/2!

How are these two questions different and how both the groups are similar in Q2 so we have to divide the result by 2!.

Answer 1

Q1. Here we have to divide 10 children into two teams A and B of 5 each. So different divisions possible = 10 C₅ = 10! / (51 *5!) 
 

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Q2. Here the players have to divide themselves into two teams of 5 each. So here any one has to lead in order to divide two teams. Now one among 10 would be selecting 4 more player for his team among rest 9 players and rest 5 player after selection of 4 would go into another team. So different divisions possible = 9 C₄ = 9! / (4! * 51) = 10! / (5! *5!).

Answer 2

divide by 2 is due to fact that team A=<1,3,5,6,7>,team B=<2,4,8,9,10>   and team A=<1,3,5,6,7>,team B=<2,4,8,9,10>  are two diffrent arragments in question 1 but in question 2 there is such ordered pair  thus divide by 2.