Color Permutations In Row Of 6 Red, 3 Blue, 3 Green Flower Pots ?

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Discussion Overview

The discussion revolves around calculating the number of color permutations for a row of 12 flower pots consisting of 6 red, 3 blue, and 3 green pots. The focus is on finding a suitable formula for this combinatorial problem.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant questions the validity of using 12! or n!/(n-r)! for this permutation problem.
  • Another suggests a method of first selecting positions for the red pots, followed by the blue pots, and so on.
  • A participant expresses a desire for a formula to avoid the lengthy process of physically determining the permutations.
  • One participant provides a general formula for permutations involving multiple types of objects, indicating how to apply it to this specific case with the given quantities of pots.

Areas of Agreement / Disagreement

There is no consensus on the correct approach or formula to use, and multiple viewpoints on how to tackle the problem are presented.

Contextual Notes

Some participants reference the general formula for permutations with multiple types of objects, but there is uncertainty regarding the specific application to the quantities given in this scenario.

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With 6 red, 3 blue and 3 green flower pots, how many color permutations in row of 12 are there ?
Its not 12! or n!/(n-r)!
 
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Why not try choosing in which places to put the red pots, then choosing which of the remaining places will have blue pots, etc...
 
I am looking for the formula for this question . It would take a long time to physically determine the answer !
For example 12! ( the number of permutations of a row of twelve jurors is 479, 001,600. How long would that take you to get 12! physically ?
One permutation of the twelve colored flower pots would be :BBRGBGRRGRRR
 
Keep in mind that for permutations with three types of objects, the general formula is:

P = \frac{n!}{k_1 ! k_2 ! k_3 !}

Where n is the total number of objects (12 balls in this case), and k_1, k_2, and k_3 are the number of each type of ball (6 red, 3 blue, and 2 green in this case). Knowing that, you should be able to get the final answer easily.
 

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