Trinomial Pyramid: an Expansion of the Ideas Related to Pascal's Triangle and the Binomial Theorem
In 2009, I gave my kids several counting problems to think about. Number 6 on the list, got me to thinking about the expansion of (a + b + c)^{n}. I realized, that the pattern would be a three diminsional equivalent to Pascal's triangle.
In Pascal's Triangle which can be used to calculate the coefficients of the expansion of (a + b)^{n}, each number is calculated by adding the two numbers directly above that number. In what I am calling the Trinomial Pyramid, each coefficient is calculated by adding the three numbers directly above it.
Also, in Pascal's triangle, you can calculate each coefficient by using the combination formula: where r is the exponent of b and n  r is the exponent of a in the expansion of (a + b)^{n} so each term has the following format: . Similarly, a term in the expansion of the trinomial: (a + b + c)^{n} is: . The program below generates the coefficients. The image below that shows how how to read the triangle.

Look at the image below of the case with n = 4 to see how to interpret the chart. Toggle the shading on and off by clicking on the gray triangle. Toggle the Pyramid's numbers for each level by clicking on the Pascal's Triangle button. The top level of the Pyramid is represented by n = 0, then n = 1 would be the level just below that, etc..
In addition to the obvious application of multiplying out a trinomial, here is another application that helped inspire me to think about this problem. Suppose you have 3 kids playing 4 rounds of a contest with 4 identical prizes for each round.  How many ways can the prizes be distributed?
 If each kid is equally likely to win a prize in any given round, what is the probability that kid a will win one prize, kid b will win no prizes, and kid c will win 3 prizes?
 Suppose you have 4 students and each one has 3 choices for an activity to do in class one day. How many ways can we distribute the students among the 3 activities?
 How do things change when we change the number of rounds to the contest along with the number of prizes or change the number of students that have 3 choices?
Answers are at the bottom of this page.



Answers:
 This is the same as the number of terms in the expansion, so there woould be 15 ways.
 Since the cell that represents the coefficient of a^{1}b^{0}c^{3} has a 4 in it, then the probability of this event is 4 divided by the total of all of the cells which is 81, so the answer is 4/81.
 Each student had 3 choices, so by the multiplication principle we have 3×3×3×3=81 total ways to distribute the 3 choices among the 4 students. Notice that this matches the sum of the coefficients mentioned in the problem above. This illustrates a three variable idea equivalent to the Binomial Theorem for two variables.
 Simply change the level of the pyramid that you are using to represent the problems. Note, that if you change the number of kids playing the prize game, or the number of choices that the students have, then this pyramid does not represent the problem. For example if we only had 2 kids or the students only had 2 choices, then we could represent the problem with Pascal's Triangle, instead of Laura's Pyramid. If we had 4 kids playing the contest or 4 choices for the students, then we would need something similar to Pascal's Triangle or Laura's Pyramid that would involve 4 variables, so it would be 4 dimensional and it would also represent the expansion of (a + b + c + d)^{n}. However, if you catch on to how to calculate all of the answers in these applications, then you can still do problems related to these without the complicated 4 dimensional image.
© 2011 Laura Shears