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Notice that teams here do not have separate identities. They are just teams (so obviously you cannot distinguish among two teams). So, if you include Boy1, Boy2 in one team then do not include them in any other team because they both are already counted as a team (doesn't matter which team).

So whatever permuatations you make, if the sequence aleady has   X,X,boy1,boy2,X,X,X,X,X,X then no need to include boy1,boy2,X,X,X,X,X,X,X,X again.

10!/(2! 2! 2! 2! 2! * 5!) = 945 ways
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