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Another Combinations Triangle?

We all know pascal's triangle, right? It's a triangle formed by adding up the two numbers above it. The values it gives form the basis for the famous formula for combinations ->

\( \dbinom{n}{k} = \frac{n!}{k!(n-k)!} \)

So what if we made another type of triangle formed by adding up the three numbers above it?

I was contemplating this and have two questions for you:

1. Using the same notation as for Pascal's Triangle , can you make a formula that will take two numbers (r, c), where r is the Row and c is the Column and output the number that appears there in this triangle?

(For example, the 7 in the picture appears at (3,3))

2. Does this triangle represent anything relating to combinations/algebra? Does it have any practical applications?

Note by Alex Li
5 months, 3 weeks ago

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2) Yes. Think about why Pascal's triangle represents coefficients of \( (x+1)^n \). The triangle that you created are the coefficients of some similar expression.

1) See above. Calvin Lin Staff · 5 months, 3 weeks ago

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@Calvin Lin @Calvin Lin. I inductively figured it was (x^2+x+1)^n but what are the applications of this? Sal Gard · 4 months, 2 weeks ago

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One of the reasons Pascal's Triangle relates to combinatorics, is that we add two adjacent numbers to get the next. For example, in the row, "14641", the first 4 represents how many ways to choose one person out of 4, and the 6 represents how many ways to choose 2 people out of 4. The first 10 in the next row represents how many ways to choose 2 people out of 5. This makes sense, because whenever we choose a combination of 5 people, either 1 or 2 of the first 4 must be chosen, and the last one is implied.

I will attempt to apply similar ideas/reasoning for this version of Pascal's Triangle. However, I do not think this relates to combinatorics because every person can be in or out, making just 2 possibilities. However, as we add 3 numbers here, there should be 3 cases. Maybe if we invent a new kind of "choose", where \(n\choose{a,b}\) is the number of ways to make a group of size \(a\) and size \(b\) from a large group of size \(n\)??? Clive Chen · 5 months, 3 weeks ago

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@Clive Chen I do not think you really need 3 different numbers in the new invented "Choose", since you can still base it off of rows and columns. It's not a 3D structure, after all. I considered this idea but it did not seem to work at all. I'm starting to think that it doesn't represent anything at all, and I really don't know how to go about approaching part 1 since any (simple) Pascal's triangle formula proof uses logic that it gets based on the results it produces. Alex Li · 4 months, 1 week ago

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@Alex Li. It appears to be the coefficients of (x^2+x+1)^n. Sal Gard · 4 months, 1 week ago

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@Sal Gard Oh, I figured that out thanks to Calvin, but I'm more curious about if these numbers could represent some sort of scenario that could be written in a sentence, like "number of ways to choose r marbles from a bag when c are identical" (though that obviously doesn't work) The answer I mostly want to know is if there is a formula to produce the numbers. Alex Li · 4 months, 1 week ago

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