Combinatoric Proof for choosing objects

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Discussion Overview

The discussion revolves around a combinatorial proof related to choosing colors for a picture, specifically addressing the equation involving binomial coefficients: nCk - (n-3)C(k-3) = (n-1)Ck + (n-2)C(k-1) + (n-3)C(k-2). The participants explore the interpretation of the left and right sides of the equation in the context of color selection, aiming to clarify their understanding of the combinatorial reasoning involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that the left side of the equation represents the total ways to choose k colors from n colors, subtracting the ways to choose k colors without including three specific colors.
  • Another participant questions the validity of the initial interpretation of (n-3)C(k-3), proposing that it should instead represent the number of ways to choose k colors while excluding one of the three specified colors.
  • Further clarification is provided on the right side of the equation, suggesting it accounts for different scenarios of color selection that include or exclude specific colors.
  • A later reply acknowledges the confusion and expresses appreciation for the clarification, indicating a shift in understanding regarding the application of the combinatorial reasoning.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the combinatorial terms involved, particularly regarding the correct application of the binomial coefficients in the context of the color problem. No consensus is reached on the initial interpretation of the left side of the equation.

Contextual Notes

There are unresolved assumptions regarding the definitions of the terms used in the combinatorial expressions and the specific conditions under which the color selections are made.

silvermane
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I've been tutoring students at my college this semester, and came across this problem with a student:

Show for all integers n\geqk\geq3,

nCk - (n-3)C(k-3) = (n-1)Ck + (n-2)C(k-1) +(n-3)C(k-2)

Since it's a combinatorial proof, I was looking at the number of ways that we could choose k colors to color a picture with, where we must have green, blue, and purple in our picture, or just 2 or 1 of them. (I want to be creative for her when I explain it)

Left Side:
Total ways to choose k colors from n colors is nCk.
The number of ways we wouldn't have them in our color choice is (n-3)C(k-3).

Right Side:
This is the side I'm having trouble figuring out. I don't know if my explanation is valid for the left side, but if anyone could help me out here, it would be great. I want to make sure I understand this for when I go to explain it to my tutee.

Thanks for your help in advance!
 
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I could be wrong, but I don't think that the color problem you chose works with n-3\choose{k-3}. This would be the number of ways in which we choose k colors while always choosing red, blue, and green- if you wanted to count the sets that don't include one of those colors (which is what it sounds like you intended), it would be n-3\choose{k}.

However, this could be the number of ways to color a picture with k colors without using all three of red, green, and blue. The right hand side could then be:
n-1\choose{k}- the number of ways to use k colors without using red

n-2\choose{k-1}- the number of ways to use k colors while using red but not using blue

n-3\choose{k-2}- the number of ways to use k colors while using red and blue but not using green
 
JThompson said:
I could be wrong, but I don't think that the color problem you chose works with n-3\choose{k-3}. This would be the number of ways in which we choose k colors while always choosing red, blue, and green- if you wanted to count the sets that don't include one of those colors (which is what it sounds like you intended), it would be n-3\choose{k}.

However, this could be the number of ways to color a picture with k colors without using all three of red, green, and blue. The right hand side could then be:
n-1\choose{k}- the number of ways to use k colors without using red

n-2\choose{k-1}- the number of ways to use k colors while using red but not using blue

n-3\choose{k-2}- the number of ways to use k colors while using red and blue but not using green

Yeah I just recently realized that it wouldn't work. That explains why I wasn't understanding how to apply my reasoning to the right-hand-side. I really like how you explained it too, if it's okay if I use this when going over it with my friend. I used to be so good at this too haha. Thank you so much for your help! :blushing:
 
No problem, and of course you can use the example.
 

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