"Concrete Mathematics"--two-term coefficients

This article begins to discuss a series of questions about the two-polynomial coefficients.

Basic identities:

The so-called two-polynomial coefficients are actually the symbols we use to represent the combination: C (n,k). It is called a two-item coefficient because it is closely related to the following two-term theorem.

Let's start with the meaning of C (n,k), which is known to represent the number of cases in which k elements are removed from n elements. By means of the combination of its characterization, we seek its calculation formula.

We take out the K elements in order, by the basic combination principle, there are N (n-1) * (n-1) ...  3*2*1, as a result of this process we have added a "certain order" condition, so on all cases should be divided by the factorial of K, to eliminate this order, that is, C (n,k) = N (n-1) * (n-2) *......*3*2*1/k! ①

Combined with some algebraic techniques, we can write it in the form of factorial, C (n,k) = n! K! * (N-K)! Ii

Further, if we put a ② right-hand n/k, we will find (n-1)!/(k-1)! * [(N-1)-(k-1)]! can be converted into another two-term coefficient C (n-1,k-1), that is, the following identity is established.

C (n,k) = N/k*c (n-1,k-1) ③

Based on the identity ③, we can prove another identity with such a subscript invariant: (n-k) C (n,k) = NC (n-1,k) ④

Before proving ④, we need to know that there is "symmetry" in the two-polynomial coefficients, i.e. C (n,k) = C (n,n-k) ⑤

This is a good understanding, from the point of view of combinatorial meaning, for C (N,k) to take k elements of the case, each one corresponds to the remainder of the N-K elements, so the number of cases is the same.

So we start ④ proof: (n-k) C (n,k) = (n-k) c (n,n-k)

=NC (n-1,n-k-1)

=NC (N-1,k)

The certificate is completed.

"Concrete Mathematics"--two-term coefficients