We shall now complicate the situation a little bit. Let us ask for the combinations of the symbols with repetitions of each symbol allowed once more in the combinations. For example, let there be only two symbols . Let us look for combinations of the form:
, , , , , , ,
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 are not-, once, twice. This is represented by the factor . Similarly, the possible choices for correspond to the factor . So, the required OGF is
On expansion, this gives :
Note that if we omit the term 1 (which corresponds to not choosing any ), the other 8 terms correspond to the 8 different combinations listed in (*). Also, observe that the exponent r of the tells us that the coefficient of 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:
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:
Each term in the expansion of this would be a power of t, and the exponent k of such a term is nothing but the total of the four contributions which went into it. The number of times a term 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 is the coefficient of in the expansion, is the answer for the above question. Further, since (*) simplifies to , it follows that the answer for the above question tallies with the coefficient specified in the following next question: calculate the coefficient of in .6
Now, consider the following problem: Express the number 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 in ]:
So, we get that coefficient of in
Let us take an example from a graphical enumeration:
A is a set V of vertices a, b, c, …, together with a set of If is considered the same as , we say the graph is . Otherwise, the graph is said to be , and we say ‘ has a direction from a to b’. The edge is called a loop. The graph is said to be of order .
If the edge-set E is allowed to be a multiset, that is, if an edge 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 and denote the numbers of undirected (respectively, directed) loopless graphs of order 5, with n edges, none of them a multiple edge, find the series and .
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 equal to 10 pairs that are available. So selection of pairs of vertices could be made in 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, .
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 and . Hence,
Thus, the counting series for labelled graphs on 5 vertices is
and the counting series for directed labelled graphs on 5 vertices is
Finally, the OGF for increasing words on an alphabet with is
The corresponding OE is which is nothing but (this explains the following problem: Verify that the number of increasing words of length 10 out of the alphabet with is the coefficient of in ).
We will continue this detailed discussion/exploration in the next article.
Until then aufwiedersehen,