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In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520

\(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way. For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two are already chosen); For the third one in 3 ways (as 4 people are already chosen); For the fourth one there is only one left.

1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520

\(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way. For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two are already chosen); For the third one in 3 ways (as 4 people are already chosen); For the fourth one there is only one left.

So we have 7*5*3*1=105

Answer: B. There is also direct formula for this:

1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

Hope it helps.

I understand how to get the answer, but I'm wondering why we assume that the order of the teams doesn't matter. The question just asks how many ways you can group them. When should we assume something matters or doesn't matter when the question doesn't specify? And also, just to clarify, when we do 8!/2!2!2!2!4!, we are also not worrying about the order within each team either right i.e. (a,b)=(b,a)?

In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520

\(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way. For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two are already chosen); For the third one in 3 ways (as 4 people are already chosen); For the fourth one there is only one left.

So we have 7*5*3*1=105

Answer: B. There is also direct formula for this:

1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

Hope it helps.

I understand how to get the answer, but I'm wondering why we assume that the order of the teams doesn't matter. The question just asks how many ways you can group them. When should we assume something matters or doesn't matter when the question doesn't specify? And also, just to clarify, when we do 8!/2!2!2!2!4!, we are also not worrying about the order within each team either right i.e. (a,b)=(b,a)?

The teams are not numbered/labeled (we don't have team #1, #2, ...), the teams are not assigned to something (for example to tournaments), ... So, the order of the teams doesn't matter.

Please check the links in my previous post for similar problems.

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14 Nov 2015, 02:37

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In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

A. 90 B. 105 C. 168 D. 420 E. 2520

Ways of choosing 2 out od a group of n people = nC2

i.e. Ways of choosing first team of 2 out of 8 = 8C2 i.e. Ways of choosing Second team of 2 out of remaining 6 = 6C2 i.e. Ways of choosing Third team of 2 out of remaining 4 = 4C2 Remaining 2 will form Forth team

Total Ways of Choosing Teams = 8C2 * 6C2 * 4C2 * 1

BUT Since the first team may come on second places and second may come on third etc. i.e.e arrangement among teams are included here which we NEED to exclude

i.e. Total Ways of Choosing Teams = 8C2 * 6C2 * 4C2 * 1 / 4! = 28*15*6/24 = 105

Answer: option B
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In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

A. 90 B. 105 C. 168 D. 420 E. 2520

The FIRST pair can be chosen in 8C2 ways

The SECOND pair can be chosen in 6C2 ways as 6 members are left after choosing 1st pair

The THIRD pair can be chosen in 4C2 ways as 4 members are left after choosing 1st two pair and so on...

Therefore total ways of splitting 8 members in 6 pairs = (8C2 * 6C2 * 4C2 * 2C2) / 4!

The entire expression is divided by 4! because the arrangements of pair have been accounted for, which need to be excluded. Arrangement for example Same pair which came at first place can be chosen at third place or forth places also while we are choosing different pairs at different places

Required answer = (28 * 15 * 6 * 1)/24 = 105

Answer: option C
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