# There are 10 students in a classroom, including the triplets Hugh, Stu, and Lou. If 3 of the 10 are randomly selected to give speeches, what is the probability that Hugh will be first, Stu will be second and Lou will be third?

$P \left(\text{Hugh first, Stu second, Lou third}\right) = \frac{1}{720} \cong .0014$

#### Explanation:

This question is asking about permutations - randomly picking elements from a population where the order of the picks matters. The general formula is:

P_(n,k)=(n!)/((n-k)!); n="population", k="picks"

We are interested in permutation involving having Hugh be first, Stu second, and Lou third. There is only 1 permutation where that can happen.

There are a total of ${P}_{10 , 3}$ permutations that can be made:

P_(10,3)=(10!)/((10-3)!)=(10!)/(7!)=(10xx9xx8xx7!)/(7!)=10xx9xx8=720

And so:

$P \left(\text{Hugh first, Stu second, Lou third}\right) = \frac{1}{720} \cong .0014$