A tourist in France wants to visit 10 different cities. If the route is randomly selected, what is the probability that she will visit the cities in alphabetical order?

$\frac{1}{3 , 628 , 800} \cong \left(2.8 \times {10}^{- 7}\right) \cong 0.00000028$

Explanation:

We have 10 different cities that can be ordered in different ways. We are looking for the probability of selecting one order in particular - alphabetical. And so the probability will be:

$\frac{1}{\text{number of permutations of 10 cities}}$

So how many permutations are there of 10 cities?

Remember that the permutation calculation for a n, population and k, number selected from that population is:

P_(n,k)=(n!)/((n-k)!) and so we have

P_(10,10)=(10!)/((10-10)!)=(10!)/(0!)=(10!)/1=10! =3,628,800

which gives the probability of doing a random tour of 10 cities in alphabetical order as:

$\frac{1}{3 , 628 , 800} \cong \left(2.8 \times {10}^{- 7}\right) \cong 0.00000028$

Let's put this a different way - the population of Connecticut in 2013 was just under $3 , 600 , 000$ people. If you had every man, woman, and child do a tour of the same 10 French cities, with each one of them doing the cities in a different order, you still wouldn't be 100% sure that a single one of them managed to do the 10 cities in order.

http://www.enchantedlearning.com/usa/states/population.shtml