Books

Let $n$ be the number of books Brilli has. We can set up an equation with the given information.

$\dbinom{n}{2}=\frac{n!}{2!(n-2)!}=\frac{n\times(n-1)\times(n-2)\times\cdots\times3\times2\times1}{2\times1\times(n-2)\times(n-3\times(n-4)\times\cdots3\times2\times1}=\frac{n\times(n-1)}{2}=190$

Multiply both sides by 2 to get

$n\times(n-1)=380$

Expanding gives us

$n^2-n=380$

Subtracting $380$ gives us a quadratic equation.

$n^2-n-380=0$

We solve by factoring.

$n^2+19n-20n-380=n\times n+19n-20n-380=n(n+19)+(-20)(n+19)=(n+(-20))(n+19)=(n-20)(n+19)=0$

We have two solutions. One of our factors needs to be 0.

$n-20=0$

$n=20$

$n+19=0$

$n=-19$

We know that Brilli can't have a negative number of books, so our only solution is $20$ . Brilli has $20$ books.

The question is asking for how many ways can $3$ books can be chosen.

$\dbinom{20}{3}=\frac{20!}{17!3!}=\frac{20\times19\times18\times\cdots\times3\times2\times1}{(17\times16\times15\times\cdots\times3\times2\times1)(3\times2\times1}$

With canceling, we have

$\dbinom{20}{3}=\frac{20\times19\times18}{3\times2\times1}=\frac{20\times19\times6\times3}{6}=20\times19\times3=\boxed{1140}$