Question

Show that, if n is a positive integer with \(n \ge 3\)\(c(n,n - 2) = \frac{{(3n - 1)}}{4}C(n,3)\) .

Short answer

It is true for all positive integers, that \(c(n,n - 2) = \frac{{(3n - 1)}}{4}C(n,3),n \ge 3\)

Step-by-step solution

 Given

The given result is \(C(n,n - 2) = \frac{{(3n - 1)}}{4}C(n,3),n \ge 3\)

The Concept of moment and forces

The objects are labelled but the boxes are unlabelled, we can use the Stirling numbers formula

\(\sum\limits_{j = 1}^k {\frac{1}{{j!}}} \sum\limits_{i = 0}^{j - 1} {{{( - 1)}^i}} C(ji){(j - i)^n}\)

Prove of result

The given result is\(C(n,n - 2) = \frac{{(3n - 1)}}{4}C(n,3),n \ge 3\)

\(C(n,n - 2)\)means when\({\rm{n}}\)people seated around\(n - 2\)empty tables, there are only two possibilities either three persons can seat in a table and other persons sitting alone or there exactly two table contains a pair of persons.

For the first possibility, 3 out of\(n\)peoples can be chosen by\(C(n,3)\)ways and they can be arranged themselves in\((3 - 1)! = 2\)! ways.

In the second possibility 4 persons can be choose by \(C(n,4)\), they can be divided into separate pair in 3 ways and among themselves only one unique arrangement of seating.

Thus, combining the both cases we obtain

\(\begin{array}{l}(n,n - 2) = 3 \times (n,4) + 2 \times (n,3)\\(n,n - 2) = (n,3) \times \left( {3 \times \frac{{n - 3}}{4} + 2} \right)\\(n,n - 2) = \frac{{3n - 1}}{4} \times (n,3){\rm{ }}\\{\rm{Hence, }}C(n,n - 2) = \frac{{(3n - 1)}}{4}C(n,3),n \ge 3\end{array}\)