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Additional Mathematics (0606) Doubt

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See the question said that TWO teams have to be choosen, so think of a two teams as one conplete event(or outcome) ,and a single team as half of an outcome,and if u know the basic counting principle it says that if two tasks are to occur in succesive than u MULTIPLY their odds, and by this the first part is 14C8 and second is 8C4*6C4
I did 8C4*6C4*2 because there were two teams and they did not specify which team has a boy or a girl. What did you get for the secirity code questions?
 
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I did 8C4*6C4*2 because there were two teams and they did not specify which team has a boy or a girl. What did you get for the secirity code questions?
I am pretty sure that you dont need to multiply by 2 as the question was based on COMBINATION not permutation where order doesnt matter, so it didnt matter which team was selected first!
 
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See the question said that TWO teams have to be choosen, so think of a two teams as one conplete event(or outcome) ,and a single team as half of an outcome,and if u know the basic counting principle it says that if two tasks are to occur in succesive than u MULTIPLY their odds, and by this the first part is 14C8 and second is 8C4*6C4

Are you sure that the first part answer for the two teams question was just 14C8, because even if you select 8 members, they can be rearranged into different teams right?
 
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I dont know I put 14C8, but then I realised it might be wrong after the exam. What did you put
 
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there are 2 ways to arrange the symbols among themselves, 6P4 ways to arrange the remaining 4 digits/letters or both and there are 5 cases in which the symbols can be next to each other. hence totally it is 5 * 6P4 * 2

What did most of the people in your class put for the 14C8 one?
 
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