Rigorous nature of combinatorics

Context: I’m a high school student, who has only ever had an introductory treatment, if that, on combinatorics. As such, the extent to which I have seen combinatoric applications is limited to situations such as “If you need a group of 2 men and 3 women and you have 8 men and 9 women, how many possible ways can you pick the group” (They do get slightly more complicated, but are usually similar).

Question: I apologise in advance for the naive question, but at an elementary level it seems as though combinatorics (and the ensuing probability that can make use of it), seems not overly rigorous. It doesn’t seem as though you can “prove” that the number of arrangements you deemed is the correct number. What if you forget a case?

I know that you could argue that you’ve considered all cases, by asking if there is another case other than the ones you’ve considered. But, that doesn’t seem to be the way other areas of mathematics is done. If I wish to prove something, I couldn’t just say “can you find a situation where the statement is incorrect” as we don’t just assume it is correct by nature.

Is combinatorics rigorous?



Combinatorics certainly can be rigourous but is not usually presented that way because doing it that way is:

  • longer (obviously)
  • less clear because the rigour can obscure the key ideas
  • boring because once you know intuitively that something works you lose interest in a rigourous argument

For example, compare the following two proofs that the binomial coefficient is n!/k!(nk)! where I will define the binomial coefficient as the number of k-element subsets of {1,,n}.

Proof 1:

Take a permutation a1,,an of n. Separate this into a1,,ak and ak+1,,an. We can permute 1,,n in n! ways and since we don’t care about the order of a1,,ak or ak+1,,an we divide by k!(nk)! for a total of n!/k!(nk)!.

Proof 2:

Let B(n,k) denote the set of k-element subsets of {1,,n}. We will show that there is a bijection


The map is defined as follows. Let πSn. Let A={π(1),π(2),,π(k)} and let B={π(k+1),,π(n)}. For each finite subset C of {1,,n} with m elements, fix a bijection gC:C{1,,m} by writting the elements of C in increasing order c1cm and mapping cii.

Define maps πA and πB on {1,,k} and {1,,nk} respectively by defining
πA(i)=gA(π(i)) and πB(j)=gB(π(j)).

We map the element πSn to the triple (A,πA,πB)B(n,k)×Sk×Snk.

Conversely, given a triple (A,σ,ρ)B(n,k)×Sk×Snk we define πSn by
π(i)={g1A(σ(i))if i{1,,k}g1B(ρ(ik))if i{k+1,,n}
where B={1,,n}A.

This defines a bijection SnB(n,k)×Sk×Snk and hence

n! = {n \choose k} k!(n – k)!

as required.


The first proof was two sentences whereas the second was some complicated mess. People with experience in combinatorics will understand the second argument is happening behind the scenes when reading the first argument. To them, the first argument is all the rigour necessary. For students it is useful to teach the second method a few times to build a level of comfort with bijective proofs. But if we tried to do all of combinatorics the second way it would take too long and there would be rioting.

Post Scriptum

I will say that a lot of combinatorics textbooks and papers do tend to be written more in the line of the second argument (i.e. rigourously). Talks and lectures tend to be more in line with the first argument. However, higher level books and papers only prove “higher level results” in this way and will simply state results that are found in lower level sources. They will also move a lot faster and not explain each step exactly.

For example, I didn’t show that the map above was a bijection, merely stated it. In a lower level book there will be a proof that the two maps compose to the identity in both ways. In a higher level book, you might just see an example of the bijection and a statement that there is a bijection in general with the assumption that the person reading through the example could construct a proof on their own.

Source : Link , Question Author : frog1944 , Answer Author : darij grinberg

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