Problem Solving With Permutations

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Discussion Overview

The discussion revolves around solving problems related to permutations and combinations, specifically focusing on scenarios involving coin tosses, ice cream cone distributions, and sports game outcomes. Participants explore the application of factorials in calculating the number of arrangements for these problems.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents three permutation problems involving coin tosses, ice cream cones, and soccer game outcomes, suggesting the use of factorials and duplicate factorials for calculations.
  • Another participant explains the reasoning behind the formula for the first problem, detailing how to arrange heads and tails using slots and permutations.
  • A later reply acknowledges a mistake in the previous explanation, indicating a correction without specifying the nature of the error.
  • Participants confirm the initial formula for the first problem and extend the discussion to the other problems, suggesting similar approaches with different denominators for the factorials.

Areas of Agreement / Disagreement

Participants generally agree on the use of factorials and the approach to solving the problems. However, there is a moment of uncertainty when one participant acknowledges a mistake, leaving some aspects of the discussion unresolved.

Contextual Notes

Some participants mention the need for different factorials in the denominators for problems with varying types of items, indicating a dependence on the specific conditions of each problem.

Who May Find This Useful

Readers interested in combinatorial mathematics, particularly those looking for assistance with permutation and combination problems in a homework context.

koolkidx45
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Hi, I have the following questions

1) A coin is tossed 9 times. In how many ways could the results be six heads and three tails?

2) A man bought two vanilla ice cream cones, three chocolate ones, four strawberry, and one butterscotch. In how many ways could he distribute them among his 10 children?

3) Juan's soccer team played a total of 14 games this season. Their record was eight wins, four losses, and two ties. In how many orders could this have happened?

I think for the above problems you take the total factorial and divide by the duplicate factorials. For ex for 1) 9! / 6! x 3! = 84

Any help is appreciated. Thanks!
 
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To try to explain the formula think about it like this. Let's say the coins were labeled 1 to 9. Let's say we were lining them up on a table and on the table we marked out three slots for tails and six slots for heads.

t t t | h h h h h h

How many ways are there to arrange coins into these slots? Well you have 9 choices for the first slot, 8 for the next, etc. So there are 9! ways.

But now we notice that for example

1 2 3 | 4 5 6 7 8 9

and

2 1 3 | 4 5 6 7 8 9

Correspond to the same coins being heads and the same set of coins being tails.

Okay so say that you have a fixed set of 3 tail-coins, and a fixed set of 6 tail coins. How many permutations does that correspond to? Well line up the tail coins first, there are 3! ways to do that and then there are 6! possibilities for the head coins.

Okay so

TOTAL_PERMUTATIONS = NUMBER_OF_POSSIBLE_SETS * PERMUTATIONS_PER_SET

9! = NUMBER_OF_POSSIBLE_SETS * (6!3!)
NUMBER_OF_POSSIBLE_SETS = (9!)/(6!3!)

So yes your formula is correct.
 
Err sorry this was wrong
 
Last edited:
welcome to pf!

hi koolkidx45! welcome to pf! :wink:
koolkidx45 said:
I think for the above problems you take the total factorial and divide by the duplicate factorials. For ex for 1) 9! / 6! x 3! = 84

yes, that's correct :smile:

(and for three types, you have three factorials on the bottom instead of two)
 
Thanks for all the help guys! So for my two other problems i would do the same thing? But the denominators would be different factorials? For ex.

3) 14!/8!*4!*2!
 
yup! :biggrin:
 

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