Hi user21820,

The following statement was given in my book under the topic combinations:

> C(n,r)=0 if r ∉ {0,1,2,3,...,n}

I understand that number of ways of choosing some x (x>n) from n items is zero. But isn't this case automatically eliminated as we already fix the domain of r to be within 0 and n? So is this meant for non-integral values of r?

A different question regarding circular permutations:

> Number of circular permutations of n things when p are alike and the rest are different, taken all at a time, distinguishing clockwise and anticlockwise arrangement is (n-1)!/p!

This formula looks similar to the linear permutation. However, is this really applicable for the circular permutation case? Further, is it possible to expand this simple formula to circular permutations where there exists more than one kind of repetitions (like 3 red balls and 4 blue balls)?

For the second question, from this answer on the main site, I came to know that one must 'master' the use of Burnside's lemma. Is there any alternate route for circular permutations with more than one kind of repetitions?

Kindly reply when you find time. Thank you.

@GuruVishnu Your question is a very good one, and actually I do not agree with your book's definition.

But first, let me clear up a misconception on your part. Whenever we define a function f, we have to specify its domain and a rule that fixes what is the output of f on each input from the domain. If you define a function C to have domain { (n,r) : n∈N ∧ r∈{0..n} }, then you cannot even talk about C(n,r) if r∉{0..n}; it is simply meaningless or forbidden.

So clearly your book wants to allow the domain of C to include pairs (n,r) such that r∉{0..n}. There is nothing wrong with choosing to do that.

However, the reason I disagree with that choice is that it is better to define C(n,k) as follows:

> C(n,0) = 1 for any n∈Z.

> C(0,k+1) = 0 for any k∈N.

> C(n+1,k+1) = C(n,k+1) + C(n,k) for any n∈Z and k∈N.

This generalizes Pascal's triangle and the combinatorial definition in your book. It would be good for you to actually construct the table of values to see how it reproduces Pascal's triangle. You can check yours against mine:

···   1  1  1  1  1  1  1  1 ···
···  −3 −2 −1  0  1  2  3  4 ···
···   6  3  1  0  0  1  3  6 ···
··· −10 −4 −1  0  0  0  1  4 ···
···  15  5  1  0  0  0  0  1 ···

It is unclear whether your book says anything about C(n,k) for n < 0, but since you didn't mention it I assume it doesn't. It would be bad to define it to be zero. Numerous theorems that involve binomial coefficients work only if it is compatible with my definition above, which means it must be nonzero for n < 0.

For example:

So from now on I'll use the definition of C that I gave. You can prove by induction that C(n,k) = 1/k! · Product { n−i : k∈{0..k−1} }. This should immediately suggest extending C(n,k) further to be defined for any real n (but still k∈N) as a polynomial function with degree k. In fact, that is a very fruitful idea. The binomial theorem (1+x)^r where r is real (not just integer) requires this extension!

For the example of circular permutations, don't even go to your question yet. First make sure you can show that there are (n−1)! ways to arrange n distinct beads on a circle (equally spaced), where we consider two ways the same iff one can be obtained from the other by rotating it around the circle.

To prove this, you must construct a bijection from the set W of ways to the set of permutations of {1..n−1}. Please try this first, and I will look at your attempt.