A few weeks ago, my friend Chris and I were bored in maths class (the origin of all good ideas) while we were studying the Binomial Theorem and Pascal's Triangle, and he randomly asked me, "What about Pascal's Pyramid?" That got me thinking about how the idea of Pascal's Triangle could be extended to different dimensions, and here are some really cool things we came up with.

##Definitions
First, Pascal's Element is simply a \(1\).

Then Pascal's Line:

\[ 1 \quad 1 \quad 1 \quad 1 \quad 1 \quad 1 \quad 1 \quad \dots\]

We define it this way because each term is the sum of the 1 terms in the column before it, and it gives some interesting patterns later on.

In [[wiki-pascals-triangle|Pascal's Triangle]], each term is the sum of the two terms above it.

Unfortunately, it is impossible to put a 3-dimensional tetrahedron onto here, so instead we will imagine Pascal's Tetrahedron layer by layer:

\[1 \\
1 \\ 1 \quad 1 \\
1 \\ 2 \quad 2 \\ 1 \quad 2 \quad 1 \\
1 \\  3 \quad 3 \\ 3 \quad 6 \quad 3 \\ 1 \quad 3 \quad 3 \quad 1 \\
1 \\ 4 \quad 4 \\ 6 \quad 12 \quad 6 \\ 4 \quad 12 \quad 12 \quad 4 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ \vdots \]

Imagine each layer being tilted forward and then stacked on top of one another. Here, each term is the sum of the 3 terms above it.

Now we move to the general definition for Pascal's Simplices (an \(n\)-simplex is a generalization of the idea of a triangle or tetrahedron to \(n\) dimensions):


##General Definition

In Pascal's \(n\)-simplex, let each element be \( \left( \begin{array}{c} d_n \\ d_{n-1} \\ d_{n-2} \\ \vdots \\ d_1 \end{array} \right) \) (read aloud as "\(d_n\) choose \(d_{n-1}\) choose \(d_{n-2}\) \(\dots\) choose \(d_1\)), where each \(d_k\) denotes the number of the \((k-1)\)-simplex where that element is found. For example, \(\displaystyle \binom{d_2}{d_1}\) is the \(d_1\)th element of the \(d_2\)th row of Pascal's triangle and is the same as the binomial coefficient "\(d_2\) choose \(d_1\)". Like Pascal's Triangle, our counting starts at 0, and for \(1 \leq k \leq (n-1)\), \(d_k\) can take any integer value \(0 \leq d_k \leq d_{k+1}\) (\(d_n\) can be any nonnegative integer). In addition, if any \(d_k < 0\), the term is 0.

Each element is the sum of the \(n\) elements in the previous (\(n-1\))-simplex; that is
\( \left( \begin{array}{c} d_n \\ d_{n-1} \\ d_{n-2} \\ \vdots \\ d_1 \end{array} \right) =  \left( \begin{array}{c} d_n-1 \\ d_{n-1}-1 \\ d_{n-2}-1 \\ \vdots \\ d_1-1 \end{array} \right) +  \left( \begin{array}{c} d_n-1 \\ d_{n-1}-1 \\ \vdots \\ d_2-1 \\ d_1 \end{array} \right) + \left( \begin{array}{c} d_n-1 \\ \vdots \\ d_3-1 \\ d_2 \\ d_1 \end{array} \right) + \dots + \left( \begin{array}{c} d_n-1 \\ d_{n-1} \\ \vdots \\ d_2 \\ d_1 \end{array} \right) \). This is called the addition identity.

##Example:
Find the value of \( \left( \begin{array}{c} 4 \\ 3 \\ 2 \end{array} \right) \).

This is the 2nd element of the 3rd row of the 4th layer of Pascal's Tetrahedron, so looking at the tetrahedron, and remembering that counting starts at 0, we see that \( \left( \begin{array}{c} 4 \\ 3 \\ 2 \end{array} \right) = 12\).

##Properties
Now, some properties of Pascal's Simplices. You will recognize many of them from Pascal's Triangle, generalized to any dimension.

The sum of the elements in any \((n-1)\)-simplex is \((d_n)^n\). This means that the sum of the elements in the \(k\)-th row of Pascal's triangle is \(2^k\), the sum of the elements in the \(k\)-th layer of Pascal's Tetrahedron is \(3^k\), and so on. This is one reason why we defined Pascal's Line to be all \(1\)'s, so that the sum of the \(k\)-th element is \(1^k\).


One way to see why this is true is the following:
Each element in Pascal's \(n\)-Simplex has \(n\) elements "below" it. It contributes to each of those elements, so it is represented \(n\) times. Thus, the sum of each \((n-1)\)-simplex is \(n\) times more than the one "above" it. Since the sum begins at \(1\), the sums are the powers of \(n\). A different proof of this will be shown later when we look at Pascal's Simplices in relation to multinomial coefficients.

Next, we present an explicit definition for each element based on binomial coefficients. This shall be called Hobbit's Theorem.

##**Hobbit's Theorem:**
\( \left( \begin{array}{c} d_n \\ d_{n-1} \\ d_{n-2} \\ \vdots \\ d_1 \end{array} \right) = \displaystyle \prod_{k=1}^{n-1} \binom{d_{k+1}}{d_k} = \displaystyle \binom{d_n}{d_{n-1}} \binom {d_{n-1}}{d_{n-2}} \binom {d_{n-2}}{d_{n-3}} \cdots \binom {d_3}{d_2} \binom {d_2}{d_1} \)

We prove this by induction on \(d_n\).
First, if \(d_n = 0\), then either

- at least 1 \(d_k\) is negative, so the entire element is 0, and at least one binomial coefficient in the product is also 0, thus Hobbit's Theorem is true.
- all \(d_k\) are 0, in which case the element is 1, being the first element in Pascal's \(n\)-Simplex. The product of the binomial coefficients is also 1 because each binomial coefficient is \( \binom{0}{0} = 1\).
Therefore, Hobbits theorem is true for all elements where \(d_n\) is 0.

Now, assume that Hobbit's theorem is true for all elements where \(d_n = k\).