How Many Ways to Split Students and Combine Outfits?

  • Context: High School 
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    Permutations
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Discussion Overview

The discussion revolves around combinatorial problems involving the arrangement of students into different groups and the calculation of outfit combinations from various clothing items. The scope includes mathematical reasoning related to combinations and permutations.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents two problems involving combinations: splitting twelve students into three groups and determining outfit combinations from clothing items.
  • Another participant proposes a formula for the outfit combinations, suggesting that the total is 36 based on the combination of shirts, jeans, and sandals.
  • A third participant provides a different approach to the first problem, using combinations to calculate the number of ways to split the students, and confirms the outfit total as 36 through a detailed breakdown of the pairing process.

Areas of Agreement / Disagreement

Participants generally agree on the total number of outfit combinations being 36, but the methods for solving the student grouping problem are presented with different approaches, and no consensus on the final answer for that problem is reached.

Contextual Notes

Some assumptions regarding the definitions of combinations and the application of formulas are not explicitly stated, and the discussion does not resolve the first problem completely.

Apophis
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:rolleyes: I can normally do combinations and permutations, but these two currently stump me. Any help is appreciated. :confused:

1) Twelve students are in a class. They are split so that five go to room A, four go to room B and three go to room C. How many different ways can this happen?

2) You have six pairs of jeans, three shirts and two pairs of sandals. How many different outfits can you wear from these choices?

:rolleyes:
 
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For number two, I think (2c1)(3c1)(6c1) describes the correct answer. There are 36 combinations. Thats evaluated as

[tex]nCk = \frac{n!}{k!(n-k)!}[/tex]
 
For The First One:

[itex](12C_5)(7C_4)(3C_3)[/itex]

For The Second One:

Take One Shirt ===> You can pair it up with 6 different jeans
And each of the above pair of shirt+jeans can be worn in two way (with 2 different pairs of sandals) Therefore total cases=

(1 x 6) x 2 =12

Similarily the above case happenes 3 times for three different shirts :

12 x 3

Ans= 36
 
Thank You

Thank You all for the help. This makes more sense to me now. :biggrin:
 

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