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Suppose teams are labelled  like $A, B, C, D, E$:

So, no. of ways $= {}^{10}C_2 \times {}^8C_2 \times {}^6C_2 \times {}^4C_2 \times {}^2C_2 \\=113400$

We can also do as $\frac{10!}{2^5} = 113400$

Now, if the teams are unlabeled, we have to divide the answer by $5! = 120$ as each of the permutation is the same.

So, required answer $= \frac{113400}{120} = 945.$

There are lot of repetitions in your answer in unlabelled case, for  example  if you select  {1,2} as first pair for one type of instances and  {3,4} as first pair for other type of instances but there can be {1,2} select as pair in remaining pair which overlaps with first type of instances.
Thanks, it was wrong. I have corrected now..
Sir can you look at my answer , I am getting different to that calculate by you using different method, can you please tell if there is any mistake.
+1 vote
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
No of ways we can select (10C2 * 2!) *(8C2 *2!) * (4C2 *2!) *(2C2 * 2!)

=3628800
that is 10! rt?

Actually one thing must be mentioned- if teams are labelled or not.
what is the meaning of labelling teams since 5 teams will be there and each will be distinct so then why to label it ,couldn't understand sir .
yes. Suppose the 5 tems were Ind, Pak, Aus, SL and SA- they are labelled. Now, in a school I ask someone to form 5 teams of 2 players each - they are unlabelled and each team is identified by its players.
Ya got it ,now if we assume it to be unlabelled then why will we divide it by 5! ?

+1 vote