# Trinomial and Multinomial theorem

1. Oct 13, 2011

### RandomMystery

I can't understand the sum notation shown in Wikipedia or in this article:

http://mathforum.org/library/drmath/view/53159.html

I want to find the sum notation for (a+b+c)^n
however I can't understand the sum notation:

I don't understand the use of brackets or what they mean here and in the binomial theorem.

I think they are what Wikipedia calls the binomial coefficient:

http://en.wikipedia.org/wiki/Binomial_theorem

I also don't understand, is that they are taking the some from i,k,j to what? Shouldn't their be a number on top of the sigma?

I would appreciate the help!

2. Oct 13, 2011

### micromass

Staff Emeritus
The notation here is a so-called multinomial coefficient, which is a generalization of a binomail coefficient. The definition is

$$\binom{n}{a,b,c}=\frac{n!}{a!b!c!}$$

A binomial coefficient

$$\binom{n}{k}$$

then equal the multinomial coefficient

$$\binom{n}{k,n-k}$$

3. Oct 13, 2011

### awkward

Adding to Micromass's remarks, the summation is meant to be over all triples i, j, k where $i \ge 0, j \ge 0, k \ge 0$ and $i+j+k=n$

4. Oct 16, 2011

### RandomMystery

Okay, thanks but I still don't understand the multi variable summation notation-

What does the i,j,k under the sigma represent and why is there nothing on "top" of the sigma?

Does this notation also assume that:

"the summation is meant to be over all triples i, j, k where i≥0,j≥0,k≥0 and
i+j+k=n"

are must this be stated separately from the Summation?

5. Oct 18, 2011

### awkward

There should be something in the text saying that the summation is over i,j,k where i+j+k=n. Sometimes you will see i+j+k=n written under the summation symbol instead.

6. Dec 3, 2011

### NguyenSP

It does'nt matter for i+j+k=n is over or under the summation symbol. It means how many ways to make i+j+k=n, where i,j,k≥ 0.
For example: (a+b+c)3
i+j+k=3
How many ways to make i+j+k=3?
3+0+0=3 => i=3,j=0,k=0
0+3+0=3 => i=0,j=3,k=0
0+0+3=3 => i=0,j=0,k=3
1+1+1=3 => i=1,j=1,k=1
2+1+0=3 => i=2,j=1,k=0
2+0+1=3 => i=2,j=0,k=1
1+2+0=3 => i=1,j=2,k=0
1+0+2=3 => i=1,j=0,k=2
0+2+1=3 => i=0,j=2,k=1
0+1+2=3 => i=0,j=1,k=2

7. Dec 4, 2011

### NguyenSP

Another way to expand (a+b+c)n is using Pascal's Triangle.