Copyright © University of Cambridge. All rights reserved.

## 'Weekly Challenge 22: Combinations of Two' printed from http://nrich.maths.org/

The diagram shows that every combination of two elements chosen
from $6$ on the line $n = 6$ corresponds to exactly one dot in the
triangular array above. This triangular array above the
dotted line in the diagram represents the triangle number $T_5$.
Conversely the diagram also shows that every dot in the triangular
array for $T_5$ corresponds to one and only one of the choices of
pairs of elements in the line $n=6$ . Hence the number of
combinations of two elements chosen from $6$ is equal to $1 + 2 + 3
+ 4 + 5$.

This generalises directly to finding the number of distinct pairs
of elements chosen from $n$ elements. Draw the triangular
array for $T_n$, that is rows of dots for $n = 1, 2, 3, ... n$. Now
in the same way as for $n=6$, every pair of elements chosen from
the bottom row can be joined by lines in the diagram to meet
in a single dot in the array for $T_{n-1}$. Conversely each
dot in $T_{n-1}$ corresponds to one and only one pair chosen from
$n$ elements on the bottom line. This shows that the number of ways
of choosing pairs from $n$ objects, generally denoted by
$^nC_2$, is $$T_{n-1} = 1 + 2 + ...+ (n-1).$$ Putting two $T_{n-1}$
triangular arrays side by side gives $n(n-1)$ dots so $$^nC_2
= T_{n-1} = \frac{1}{2}n (n - 1).$$