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combinatorics question (1 Viewer)

kenkap

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In how many ways can a group of twelve people be divided into three groups of four to play with each other??

Explanation will be appreciated...thanks!
 

GoldyOrNugget

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I think it's

(12C4 * 8C4) / 3!

Choose 4 from the initial 12, then 4 from the remaining 8, then the remaining 4 are their own group. Then, to make the ordering in which you chose the groups unimportant, divide by 3!.

EDIT: this is easier to see for a smaller case. Consider picking 2 groups of 2 from 4. The people are ABCD.

The solution would be 4C2 / 2!. The 4C2 comes from the two people you pick for the first group. The options would seem to be

(AB, CD)
(AC, BD)
(AD, BC)
(BC, AD)
(BD, AC)
(CD, AB)

but note that (AB, CD) is the same as (CD, BA), and there are other duplicate sets. Dividing by 2! fixes this issue by removing sets of groups that are the same.
 
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kenkap

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I think it's

(12C4 * 8C4) / 3!

Choose 4 from the initial 12, then 4 from the remaining 8, then the remaining 4 are their own group. Then, to make the ordering in which you chose the groups unimportant, divide by 3!.
mate it says 3 groups of four, not four groups of three....u got tricked!!
 

simskee

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Is it
(12C4 * 8C4 * 4C4 )/3!?
 
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