Vikalpa, is a Sanskrit/Hindi word, often used in context of choice making. The ancient Indian Scripture, Sthananga Sutra, uses this word as a topic name for a text on Permutation and Combination. For the study of subject matter such as, “Vikalpa“, I will like to use the term “Combinatorics“, mainly because the terms, Permutation, Combination are Mathematical quantities, and the word Combinatorics captures the wider essence of Mathematics of “choice making”. So whenever we deal with problems which involve enumerating possible choices, often conforming certain patterns, we will be doing Combinatorics. Combinatorial problems are the theme of many of the every day calculations, for example, the possible decisions to traverse from one place to other place. And unless some patterns are not discovered, calculations can become pretty tedious. But when patterns are found, and effectively used, things can become fun. A very fine approach for Combinatorial proofs are given in Proofs that really count: The art of Combinatorial proof, by Arthur T. Benjamin and Jennifer J. Quinn.

Consider this question — “How many ways persons can be seated across a round table having chairs around it, where only relative order of persons matter?”. This question is a part of the problem statement of the plan for seating people across a round table at dinner, where situation like who should/shouldn’t sit adjacent to a person are often dealt. If we consider a person as an element of set,

, we need all possible arrangements

,

where is a bijective mapping on and into the set and arrangements are considered equivalent when,

, being an integer constant.

This is like finding all arrangements of persons, where a given arrangement should not be circular shift of another one. Since for a given arrangement there are number of circular shifts, so total number of ways in which people can be seated across a round table, is , i.e., . So when on a clear sky night, if we want to sketch straight lines between a given number of stars in such a way that all stars are on the final sketch and they are crossed only once, then we have number of possible sketches. What if we want to make , such sketches( each sketch should have atleast one star on it ), which do not have any star in common, when we have stars? Or, suppose if we have number of identical round tables, how many ways are there to seat people across them, in such a way that at least one person should seat across a round table ?

If there are , number of stars in th cycle, total number of solutions is given by,

,

where the summation ranges on all solutions of the equation . The factor , is number of ways of choosing groups of stars out of stars. The factor , is number of sketches that can be made on number of stars ( the same total number of seating arrangement solution ). Also, since solutions of can be permutation of each other, so I provided an expression in denominator. A simple reduction can be acheived by expanding the combinations–

.

Like the way there is no explicit formula for factorial, permutation and combination; so is the situation with the above solution, and in literature we use as their symbolic representation. This quantity, , is called unsigned Stirling number of first kind. We will now explore falling factorial power, , to gain more insight into Stirling number of first kind:-

.

Unsigned coefficient of in the above product equals:-

Unsigned Coefficient of , in

,

where are distinct number chosen from the set . The summation thus runs for cases which is equal to total number of solutions of , . Literatures assert that, unsigned coefficient of , in is . Does this mean,

,

where are distinct numbers from the set while the summations on both sides are running for equal number of time ? Remember, ’s can have same values.

I am unable to think now, and will work out proof of last equality in few days. Going to sleep. Keep thinking and reading,

Mukul

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