how many ways can we form a team of 2 men and 2 women such that no couple is a part of team ?

Question

There are 10 couples at party .In this question I have only doubt that if I chose 2 husbands first ,now for wives remaining choice will be 8 so it will be 10C2 *8C2 ,Now I could have done it other way around like first chosing 2 wives and then removing their corresponding husbands so again no of ways in which team can be formed is 10C2*8C2 ,now why can't we add these two cases ?

Comments

because both are the same- you will be double counting the possibilities.

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sir , even I was thinking the same but then I thought that in one case I am considering women and in another case I am considering men so I got confused at this point and still I am confused only .

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the order in the team doesn't matter. So, even if we choose men first and then women or vice verse, they are the same.

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Order of choosing combinations of them here doesn't matter because it will give you duplicate values and that is why you can not take it twice.