Can anyone help me about this traingle question? I am confused

Can anyone please help me about this triangle question? I understood why we did 6 * 5* 4 but I didn't understand why de divide that by 3! . I also don't understand to find the number of ways why we multiply and why don't we add. Thank you for help please help me.

Hi @fabulousgrizzly!
Thanks for reaching out  we're happy to help!To begin with, let's start from your first question about \(3!\)
(for better understanding let's numerate lines as line 1, line 2, line 3, line 4, line 5 and line 6)
First, we counted such a strange number \(6\cdot 5\cdot 4.\) What does this number mean? By this number we counted all possible triples of lines, which means that the triples {line 1, line 2, line 3} and {line 3, line 1, line 2} are counted as different ones. But triple {line 1, line 2, line 3} and triple {line 3, line 1, line 2} determine the same triangle (which appears because of intersections of lines 1, 2 and 3). So we want to count only one of the similar triples, for example we want to count only {line 1, line 2, line 3} triple, but not {line 3, line 1, line 2}.
Now, when we decided to eliminate all unnecessary triples from the groups of similar ones, we want to count, how many similar triples can appear with 3 different lines. Let's use the same 3 lines: line 1, line 2 and line 3.
So we have 3 groups, in each group two smaller groups and in each small group  one "element" (a triple). To count the number of triples it is enough to count the number of "elements": \(3\text{ groups}\times 2\text{ groups in each}\times 1\text{ element in each}=3!\)
So by number \(6\cdot 5\cdot 4\) we counted:
$$\text{All triples in general}=\text{all different triples}\times\text{number of similar triples to each different one}$$ $$6\cdot 5\cdot 4 = \text{all different triples}\times3!$$
So, $$\text{all different triples} = \frac{6\cdot 5\cdot 4}{3!}$$
We can use the same idea from above in order to count the number of ways to study 3 topics out of 5:
You have 5 groups, every group of which contains 4 smaller groups, with each smaller group containing 3 "ways" each. Here each "way" means a triple of subjects, where the order of the subjects is important.
So the total number of "ways" will be: \( \left( 5\text{ groups} \right) \times \left( 4\text{ groups in each} \right) \times \left( 3\text{ element in each} \right) =5\cdot 4\cdot 3.\)
To find the number of the triples of subjects where the order of the subjects doesn't matter, we have to divide by \(\left( \text{number of similar triples}\right), \) which is \(3!\)
So the answer is equal to \(\frac{5\cdot 4\cdot 3}{3!}=\frac{60}{6}=\boxed{10}.\) 
@fabulousgrizzly Here's another forum explanation for why we have to divide by 6.