PD posted this wonderful combinatorial puzzle. It was so nicely knit, really enjoyed it. Here it is :
If you have 2 numbers, if you order them using = and < there are three possible orderings
- a < b, a = b , b < a
for three numbers, this is 13
a < b < c
a = b < c
b < a < c
b < a = c
b < c < a
a < c < b
a = c < b
c < a < b
c < a = b
c < a < b
a < b = c
b = c < a
a = b = c
for four it turns out be 75.
the problem is to characterize such a number and to devise a way to calculate the number of
possible orderings with n elements.
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Two approaches to solve it is here. Let me know when you find more.
