Trinomial triangle

The trinomial triangle is a variation of Pascal's triangle. The difference between the two is that an entry in the trinomial triangle is the sum of the three (rather than the two in Pascal's triangle) entries above it:

${\begin{matrix}&&&&1\\&&&1&1&1\\&&1&2&3&2&1\\&1&3&6&7&6&3&1\\1&4&10&16&19&16&10&4&1\end{matrix}}$

The $k$-th entry of the $n$-th row is denoted by

${n \choose k}_{2}$.

Rows are counted starting from 0. The entries of the $n$-th row are indexed starting with $-n$ from the left, and the middle entry has index 0. The symmetry of the entries of a row about the middle entry is expressed by the relationship

${n \choose k}_{2}={n \choose -k}_{2}$

Properties

The $n$-th row corresponds to the coefficients in the polynomial expansion of the expansion of the trinomial $(1+x+x^{2})$ raised to the $n$-th power:

$\left(1+x+x^{2}\right)^{n}=\sum _{j=0}^{2n}{n \choose j-n}_{2}x^{j}=\sum _{k=-n}^{n}{n \choose k}_{2}x^{n+k}$

or, symmetrically,

$\left(1+x+1/x\right)^{n}=\sum _{k=-n}^{n}{n \choose k}_{2}x^{k}$,

hence the alternative name trinomial coefficients because of their relationship to the multinomial coefficients:

${n \choose k}_{2}=\sum _{\textstyle {0\leq \mu ,\nu \leq n \atop \mu +2\nu =n+k}}{\frac {n!}{\mu !\,\nu !\,(n-\mu -\nu )!}}$

Furthermore, the diagonals have interesting properties, such as their relationship to the triangular numbers.

The sum of the elements of $n$-th row is $3^{n}$.

Recurrence formula

The trinomial coefficients can be generated using the following recurrence formula:

${0 \choose 0}_{2}=1$,

${n+1 \choose k}_{2}={n \choose k-1}_{2}+{n \choose k}_{2}+{n \choose k+1}_{2}$ for $n\geq 0$,

where ${n \choose k}_{2}=0$ for $\ k<-n$ and $\ k>n$.

Central trinomial coefficients

The middle entries of the trinomial triangle

1, 1, 3, 7, 19, 51, 141, 393, 1107, 3139, … (sequence A002426 in the OEIS)

were studied by Euler and are known as central trinomial coefficients.

The $n$-th central trinomial coefficient is given by

${n \choose 0}_{2}=\sum _{k=0}^{n}{\frac {n(n-1)\cdots (n-2k+1)}{(k!)^{2}}}=\sum _{k=0}^{n}{n \choose 2k}{2k \choose k}.$

Their generating function is

$1+x+3x^{2}+7x^{3}+19x^{4}+\ldots ={\frac {1}{\sqrt {(1+x)(1-3x)}}}.$

Euler noted the following exemplum memorabile inductionis fallacis ("notable example of fallacious induction"):

$3{n+1 \choose 0}_{2}-{n+2 \choose 0}_{2}=F_{n}(F_{n}+1)$ for $0\leq n\leq 7$,

where $F_{n}$ is the n-th Fibonacci number. For larger $n$, however, this relationship is incorrect. George Andrews explained this fallacy using the general identity

$2\sum _{k\in \mathbb {Z} }\left[{n+1 \choose 10k}_{2}-{n+1 \choose 10k+1}_{2}\right]=F_{n}(F_{n}+1).$

Applications

In chess

Number of ways to reach a cell with the minimum number of moves

The triangle corresponds to the number of possible paths that can be taken by the king in a game of chess. The entry in a cell represents the number of different paths (using a minimum number of moves) the king can take to reach the cell.

In combinatorics

The coefficient of $x^{k}$ in the polynomial expansion of $\left(1+x+x^{2}\right)^{n}$ specifies the number of different ways of randomly drawing $k$ cards from two sets of $n$ identical playing cards. For example, in such a card game with two sets of the three cards A, B, C, the choices look like this:

Number of selected cards	Number of options	Options
0	1
1	3	A, B, C
2	6	AA, AB, AC, BB, BC, CC
3	7	AAB, AAC, ABB, ABC, ACC, BBC, BCC
4	6	AABB, AABC, AACC, ABBC, ABCC, BBCC
5	3	AABBC, AABCC, ABBCC
6	1	AABBCC

In particular, this results in ${24 \choose 12-24}_{2}={24 \choose -12}_{2}={24 \choose 12}_{2}$ as the number of different hands in a game of Doppelkopf.

Alternatively, it is also possible to arrive at this number by considering the number of ways of choosing $p$ pairs of identical cards from the two sets, which is ${n \choose p}$. The remaining $k-2p$ cards can then be chosen in ${n-p \choose k-2p}$ ways, which can be written in terms of the binomial coefficients as

${n \choose k-n}_{2}=\sum _{p=\max(0,k-n)}^{\min(n,[k/2])}{n \choose p}{n-p \choose k-2p}$.

For example,

$6={3 \choose 2-3}_{2}={3 \choose 0}{3 \choose 2}+{3 \choose 1}{2 \choose 0}=1\cdot 3+3\cdot 1$.

The example above corresponds to the three ways of selecting two cards without pairs of identical cards (AB, AC, BC) and the three ways of selecting a pair of identical cards (AA, BB, CC).

Further reading

• Leonhard Euler (1767). "Observationes analyticae ("Analytical observations")". Novi Commentarii academiae scientiarum Petropolitanae. 11: 124–143.