# The trinomial theorem

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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°.