Three integers are randomly chosen from $1$ to $5$ with repetition not allowed. Find the probability that the integers chosen can represent the lengths of each side of a triangle
I first found the number of ways to choose the three integers by $\binom{5}{3} = 10$ ways to choose three side lengths from the numbers $1$ through $5$.
Here's where I'm stuck. There are 10 combinations:
$${1,2,3}$$ $${1,2,4}$$ $${1,2,5}$$ $${1,3,4}$$ $${1,3,5}$$ $${1,4,5}$$ $${2,3,4}$$ $${2,3,5}$$ $${2,4,5}$$ $${3,4,5}$$
But now how do I know which combinations can represent the lengths of a triangle
It is known that there is a triangle with sides $a$, $b$, $c$ if and only if three triangle inequalities hold: $$\begin{cases}a+b>c\\b+c>a\\ c+a>b\end{cases}$$ In our case repetition is not allowed, so the numbers $a$, $b$, $c$ are different, and we can arrange them: $$a< b<c.$$ And we only need to check if $$a+b>c.$$
You found all the triples. There are $\binom53=10$ of them. Only $3$ of them fit: $$2,3,4$$ $$2,4,5$$ $$3,4,5$$
All the others don’t fit. The triples of the form $1,b,c$ aren’t good, since always $1+b\le c$. And the triple $2,3,5$ is also bad: $2+3=5$.