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zura
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50% (05:30) correct 50% (00:19) wrong based on 2 sessions

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1. There are 6 students, and you want to form 3 groups of 2 each. How many groups are possible?

2. There are 6 students, and they are chosen to form 3 groups to make presentation. Each group will be assigned Asia, Africa or America as their presentation subject. And the question is how many groups could be formed.

3. How many ways can you split 14 people into 7 pairs?



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oldstudent
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The above questions can be solved by the sub-group rule also discussed by Bunuel in many posts:
Check this out: combination-55369.html

Please also keep the answer options while posting any questions in the forum.
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sudhanshushankerjha
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1) 6!/( 2! *2! *2!)

3) 14! /(2!)^7

2)I m not sure..bt i think its ans is 3^3
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1) 6!/( 2! *2! *2!)

2) 6!/(( 2! *2! *2!)*3!)
Need to divide by 3! as the order of the group matters.

3) 14! /(2!)^7
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1. There are 6 students, and you want to form 3 groups of 2 each. How many groups are possible?

Answer: 6!/(2!*2!*2!*3!)

-There 6! ways to arrange 6 students to 6 positions,
-There 2! to arrange 2 students in each team in each of 3 teams- the order within teams does not matter.
-there 3! ways to arrange 3 teams-the order of the teams does not matter

2. There are 6 students, and they are chosen to form 3 groups to make presentation. Each group will be assigned Asia, Africa or America as their presentation subject. And the question is how many groups could be formed.

(6!*/(2!*2!*2!*3!))*3! - the same as in one but need to be multiplied by 3! because there are 3! different ways to asssign 3 different topics to 3 different teams.
So, its 6!/(2!*2!*2!)

3. How many ways can you split 14 people into 7 pairs?

14!/(2!^7*7!)

Needs to be divided by 7! because the order of 7 pairs does not matter, and there are 7! different arangements possible to assign 7 divverent teams to 7 different spots.



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Bunuel
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