A Primer: Generating Functions: Part II: for RMO/INMO 2019

We shall now complicate the situation a little bit. Let us ask for the combinations of the symbols \alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} with repetitions of each symbol allowed once more in the combinations. For example, let there be only two symbols \alpha_{1}, \alpha_{2}. Let us look for combinations of the form:

\alpha_{1}, \alpha_{2}, \alpha_{1}\alpha_{2}, \alpha_{1}\alpha_{1}, \alpha_{2}\alpha_{2}, \alpha_{1}\alpha_{1}\alpha_{2}, \alpha_{1}\alpha_{2}\alpha_{2}, \alpha_{1}\alpha_{1}\alpha_{2}\alpha_{2}

where, in each combination, each symbol may occur once, twice, or not at all. The OGF for this can be constructed by reasoning as follows: the choices for \alpha_{1} are not-\alpha_{1}, \alpha_{1} once, \alpha_{1} twice. This is represented by the factor (1+\alpha_{1}t+\alpha_{1}^{2}t^{2}). Similarly, the possible choices for \alpha_{2} correspond to the factor (1+\alpha_{2}t+\alpha_{2}^{2}t^{2}). So, the required OGF is (1+\alpha_{1}t+\alpha_{1}^{2}t)(1+\alpha_{2}t+\alpha_{2}^{2}t^{2})

On expansion, this gives : 1+(\alpha_{1}+\alpha_{2})t+(\alpha_{1}\alpha_{2}+\alpha_{1}^{2}+\alpha_{2}^{2})t^{2}+(\alpha_{1}^{2}\alpha_{2}+\alpha_{1}\alpha_{2}^{2})t^{3}+(\alpha_{1}^{2}\alpha_{2}^{2})t^{4}

Note that if we omit the term 1 (which corresponds to not choosing any \alpha), the other 8 terms correspond to the 8 different combinations listed in (*). Also, observe that the exponent r of the t^{r} tells us that the coefficient of t^{r} has the list or inventory of the r-combinations (under the required specification — in this case, with the restriction on repetitions of symbols) in it:


In the light of the foregoing discussion, let us now take up the following question again: in how many ways, can a total of 16 be obtained by rolling 4 dice once?; the contribution of each die to the total is either a “1” or a “2” or a “3” or a “4” or a “5” or a “6”. The contributions from each of the 4 dice have to be added to get the total — in this case, 16. So, if we write: t^{1}+t^{2}+t^{3}+t^{4}+t^{5}+t^{6}

as the factor corresponding to the first die, the factors corresponding to the other three dice are exactly the same. The product of these factors would be:

(*) (t+t^{2}+t^{3}+t^{4}+t^{5}+t^{6})^{4}

Each term in the expansion of this would be a power of t, and the exponent k of such a term t^{k} is nothing but the total of the four contributions which went into it. The number of times a term t^{k} can be obtained is exactly the number of times k can be obtained as a total on a throw of the four dice. So, if \alpha_{k} is the coefficient of t^{k} in the expansion, \alpha_{16} is the answer for the above question. Further, since (*) simplifies to (\frac{t(1-t^{6})}{1-t})^{4}, it follows that the answer for the above question tallies with the coefficient specified in the following next question: calculate the coefficient of t^{12} in (\frac{(1-t^{6})}{(1-t)})^{4}.6

Now, consider the following problem: Express the number N(n,p) of ways of obtaining a total of n by rolling p dice, as a certain coefficient in a suitable product of binomial expansions in powers of t. [ this in turn, is related to the observation that the number of ways a total of 16 can be obtained by rolling 4 dice once is the same as the coefficient of t^{12} in (\frac{1-t^{6}}{1-t})^{4}]:

So, we get that N(n,p)= coefficient of t^{n-p} in (\frac{1-t^{6}}{1-t})^{p}

Let us take an example from a graphical enumeration:

A \it {graph} G=G(V,F) is a set V of vertices a, b, c, …, together with a set E=V \times V of \it {edges} (a,b), (a,a), (b,a), (c,b), \ldots If (x,y) is considered the same as (y,x), we say the graph is \it{undirected}. Otherwise, the graph is said to be \it{directed}, and we say ‘(a,b) has a direction from a to b’. The edge (x,x) is called a loop. The graph is said to be of order |V|.

If the edge-set E is allowed to be a multiset, that is, if an edge (a,b) is allowed to occur more than once, (and, this may be called a ‘multiple edge’), we refer to the graph as a general graph.

If \phi_{5}(n) and \psi_{5}(n) denote the numbers of undirected (respectively, directed) loopless graphs of order 5, with n edges, none of them a multiple edge, find the series \sum \phi_{5}(n)t^{n} and \sum \psi_{5}(n)t^{n}.

Applying our recently developed techniques to the above question, a graph of 5 specified vertices is uniquely determined once you specify which pairs of vertices are ‘joined’. Suppose we are required to consider only graphs with 4 edges. This would need four pairs of vertices to be selected out of the total of 5 \choose 2 equal to 10 pairs that are available. So selection of pairs of vertices could be made in 10 \choose 4 ways. Each such selection corresponds to one unique graph, with the selected pairs being considered as edges. More informally, having selected a certain pairs of vertices, imagine that the vertices are represented by dots in a diagram and join the vertices of each selected pair by a running line. Then, the “graph” becomes a “visible” object. Note that the number of graphs is just the number of selections of pairs of vertices. Hence, \phi_{5}(4)=10 \choose 4.

Or, one could approach this problem in a different way. Imagine that you have a complete graph on 5 vertices — the “completeness” here means that every possible pair of vertices has been joined by an edge. From the complete graph which has 10 edges, one has to choose 4 edges — any four, for that matter — in order to get a graph as required by the problem.

On the same lines for a directed graph, one has a universe of 10 by 2, that is, 29 edges to choose from, for, each pair x,y gives rise to two possible edges (x,y) and (y,x). Hence,

\psi_{5}(4)=20 \choose 4.

Thus, the counting series for labelled graphs on 5 vertices is 1 + \sum_{p=1}^{10} {10 \choose p}t^{p}
and the counting series for directed labelled graphs on 5 vertices is
1+ \sum_{p=1}^{20}{20 \choose p}t^{p}.

Finally, the OGF for increasing words on an alphabet {a,b,c,d,e} with a<b<c<d<e is

(1+at+a^{2}t^{2}+\ldots)(1+bt+b^{2}t^{2}+\ldots)(1+ct+c^{2}t^{2}+\ldots)\times (1+dt+d^{2}t^{2}+\ldots)(1+et+e^{2}t^{2}+\ldots)

The corresponding OE is (1+t+t^{2}+t^{3}+\ldots)^{5} which is nothing but (1-t)^{-5} (this explains the following problem: Verify that the number of increasing words of length 10 out of the alphabet \{a,b,c,d,e \} with a<b<c<d<e is the coefficient of t^{10} in (1-t)^{-5} ).

We will continue this detailed discussion/exploration in the next article.

Until then aufwiedersehen,
Nalin Pithwa

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