Win the Coin Toss Game at the Fair - Challenge Problem

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Discussion Overview

The discussion revolves around a coin toss game where participants explore how to achieve a total of 10 using four coins that can each yield values from 0 to 5. The focus is on identifying the different combinations that can sum to 10, including considerations of arrangement and repetition of values.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Marci presents the problem of finding combinations of four coins that add up to 10, with possible values of 0 to 5.
  • Some participants list various combinations that sum to 10, including examples like 5+5+0+0 and 5+4+1+0.
  • One participant emphasizes the importance of considering distinct arrangements of combinations, suggesting that arrangements of unique values should be calculated.
  • Another participant proposes that there are 13 unique combinations, each of which can be arranged in multiple ways, initially suggesting a calculation of 4! times the number of unique combinations.
  • A later reply corrects the initial arrangement calculation by factoring in repetitions, leading to a revised total of 200 ways to achieve the sum of 10.

Areas of Agreement / Disagreement

Participants generally agree on the need to consider both unique combinations and their arrangements, but there is no consensus on the final count of combinations due to differing approaches to handling repetitions.

Contextual Notes

Some assumptions about the counting method and the treatment of repeated values are not fully resolved, leading to different interpretations of the total number of combinations.

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Marci is working a coin toss game at the fair. You get 4 coins to toss to try to get numbers that add up to 10. You can get a
0,1,2,3,4 or 5. How many different combinations of 10 are there? (Hint: 5+5+0+0 is different from 5+0+5+0)
 
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Combinations

Marci is working a coin toss game at the fair. You get 4 coins to toss to try to get numbers that add up to 10. You can get a 0,1,2,3,4 or 5, How many different combinations of 10 are there? (Hint 5+5+0+0 is different than 5+0+5+0)
 
Last edited:
Hello and welcome to MHB, Ann and aditup! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?
 
I started with the combination of how many different ways you can add 4 coins to get 10.

5+5+0+0
5+4+1+0
5+3+2+0
5+2+2+1
4+4+2+0
4+3+2+1
4+3+3+0
3+3+3+1
3+2+3+2
 
aditup said:
I started with the combination of how many different ways you can add 4 coins to get 10.

5+5+0+0
5+4+1+0
5+3+2+0
5+2+2+1
4+4+2+0
4+3+2+1
4+3+3+0
3+3+3+1
3+2+3+2

Okay...good! Now, let's first look at those combinations having 4 distinct values:

5+4+1+0
5+3+2+0
4+3+2+1

How many ways can we arrange each?
 
Ann said:
Marci is working a coin toss game at the fair. You get 4 coins to toss to try to get numbers that add up to 10. You can get a
0,1,2,3,4 or 5. How many different combinations of 10 are there? (Hint: 5+5+0+0 is different from 5+0+5+0)

Well... There are plenty of ways to get 10 by adding 4 numbers from 0-5, let's see them:

0055 1315 2224
0145 1225 2233
0235 1144 2242
0244 1234
0334 1333
(continuing would simply mean repeating the same addition, so we're not going to do that.

I BELIEVE that's all of the UNIQUE ways of doing it.

So there are 13 unique ways, each with 4! ways of arranging them, so it'd be $$4! \times\ 13$$.
 
IHateFactorial said:
Well... There are plenty of ways to get 10 by adding 4 numbers from 0-5, let's see them:

0055 1315 2224
0145 1225 2233
0235 1144 2242
0244 1234
0334 1333
(continuing would simply mean repeating the same addition, so we're not going to do that.

I BELIEVE that's all of the UNIQUE ways of doing it.

So there are 13 unique ways, each with 4! ways of arranging them, so it'd be $$4! \times\ 13$$.

Sorry, my bad. I forgot to take into account the repeated numbers in each unique combination.

I.e. 1333 can only be cominated in 4 ways, no 4! ways (they'd be: 1333, 3133, 3313, 3331).

I'll fix that in a bit.

Factoring in for repeated numbers we have
$$ (4 \cdot 4!) + (4 \cdot 4!/2!) + (3 \cdot 4!/2!2!) + (2 \cdot 4!/3!) = 96 + 48 + 18 + 8 = 200$$
ways of getting 10.
 
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