The answer is the number of subsets with cardinality 3 that we can extract from a set with cardinality 31.
The answer is the binomial coefficient
[tex]\displaystyle\binom{n}{k}=\dfrac{n!}{k!(n-k)!}[/tex]
Where the factorial is defined as
[tex]n!=n(n-1)(n-2)(n-3)\ldots 3\cdot 2[/tex]
So, your answer is
[tex]\displaystyle\binom{31}{3}=\dfrac{31!}{3!\;28!}=\dfrac{31\cdot 30\cdot 29\cdot 28\ldots 2}{3\cdot 2\cdot 28\ldots 2}=\dfrac{31\cdot 30\cdot 29}{3\cdot 2}=31\cdot29\cdot5=4495[/tex]
