## Thursday, 7 March 2013

### THURSDAY, 7 MARCH 2013

Today is the $66^{th}$ day of the year.

$66 = 2 \times 3 \times 11$

$66 = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 8 + 9 + 10 + 11$ which means that $66$ is the eleventh triangular number, see A000217. The $n^{th}$ triangular number is calculated by the formula $\frac {n(n+1)}{2}$.

If we consider the number of ways that two different numbers can be selected from the set {0, 1, 2, ..., 11} where (a,a) is allowed but (a, b) is considered the same as (b, a) then we might come up with an array like this where X identifies a member of the set

1  2  3  4  5  6  7  8  9 10 11
1  X  X  X  X  X  X  X  X  X  X  X
2     X  X  X  X  X  X  X  X  X  X
3        X  X  X  X  X  X  X  X  X
4           X  X  X  X  X  X  X  X
5              X  X  X  X  X  X  X
6                 X  X  X  X  X  X
7                    X  X  X  X  X
8                       X  X  X  X
9                          X  X  X
10                             X  X
11                                X

Clearly, there are $11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1$ crosses and we already know that this adds up to $66$.
However, this argument could be applied to any number not just $11$ so we easily come to the conclusion that
The number of ways two different numbers can be selected from the set {1,2,...,n} with repetition is the $n^{th}$ triangular number or $\frac {n(n+1)}{2}$.