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