When I was teaching the binomial theorem (or, to be more accurate, the binomial
*expansion*) to my long-suffering students, one of them asked me if there was a
*trinomial* theorem. Well, of course there is, although in fact expanding sums
of greater than two terms is generally not classed as a theorem described by the
number of terms. The general result is

\[ (x_1+x_2+\cdots+x_k)^n=\sum_{a_1+a_2+\cdots+a_k=n} {n\choose a_1,a_2,\ldots,a_k}x_1^{a_1}x_2^{a_2}\cdots x_k^{a_k} \]

so in particular a “trinomial theorem” would be

\[ (x+y+z)^n=\sum_{a+b+c=n}{n\choose a,b,c}x^ay^bz^c. \]

Here we define

\[ {n\choose a,b,c}=\frac{n!}{a!b!c!} \]

and this is known as a *trinomial coefficient*; more generally, for an arbitrary
number of variables, it is a *multinomial coefficient*. It is guaranteed to be
an integer if the lower values sum to the upper value.

So to compute \((x+y+z)^5\) we could list all integers \(a,b,c\) with \(0\le a,b,c\le 5\) for which \(a+b+c=5\), and put them all into the above sum.

But of course there’s a better way, and it comes from expanding \((x+y+z)^5\) as a binomial \((x+(y+z))^5\) so that

\begin{array}{rcl}
(x+(y+x))^5&=&x^5\\

&&+5x^4(y+z)\\

&&+10x^3(y+z)^2\\

&&+10x^2(y+z)^3\\

&&+5x(y+z)^4\\

&&+(y+z)^5
\end{array}

Now we can expand each of those binomial powers:

\begin{array}{rcl}
(x+(y+x))^5&=&x^5\\

&&+5x^4(y+z)\\

&&+10x^3(y^2+2yz+z^2)\\

&&+10x^2(y^3+3y^2z+3yz^2+z^3)\\

&&+5x(y^4+4y^3z+6y^2z^2+4yz^3+z^4)\\

&&+(y^5+5y^4z+10y^3z^2+10y^2z^3+5yz^4+z^5)
\end{array}

Expanding this produces

\begin{split}
x^5&+5x^4y+5x^4z+10x^3y^2+20x^3yz+10x^3z^2+10x^2y^3+30x^2y^2z+30x^2yz^3\\

&+10x^2z^3+5zy^4+20xy^3z+30xy^2z^2+20xyz^3+5xz^4+y^5+5y^4z+10y^3z^2\\

&+10y^2z^3+5yz^4+z^5
\end{split}

which is an equation of rare beauty.

But there’s a nice way of setting this up, which involves writing down Pascal’s triangle to the fifth row, and putting a fifth row, as a column, on the side. Then multiply across:

\begin{array}{lcccccccccc}
1&&&&&&1&&&&&\\

5&&&&&1&&1&&&&\\

10\quad\times&&&&1&&2&&1&&&\\

10&&&1&&3&&3&&1&&\\

5&&1&&4&&6&&4&&1&\\

1&1&&5&&10&&10&&5&&1
\end{array}

to produce the final array of coefficients (with index numbers at the left):

\begin{array}{l*{10}{c}}
0\qquad{}&&&&&&1&&&&&\\

1&&&&&5&&5&&&&\\

2&&&&10&&20&&10&&&\\

3&&&10&&30&&30&&10&&\\

4&&5&&20&&30&&20&&5&\\

5&1&&5&&10&&10&&5&&1
\end{array}

Row \(i\) of this array corresponds to \(x^{5-i}\) and all combinations of powers \(y^bz^c\) for \(0\le b,c\le i\). Thus for example the fourth row down, corresponding to \( i=3 \), may be considered as the coefficients of the terms

\[ x^2y^3,\quad x^2y^2z,\quad x^2yz^2,\quad xz^3. \]

Note that the triangle of coefficients is symmetrical along all three centre lines, as well as rotationally symmetric by 120°.

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