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I have an excersie that says:

¿How many triangles can be formed, with the vertex of a hexagon?

I think this:

$1$) Ok, it is requesting triangles, therefore, it is not permutation since it takes sets of 3 vertex.

Now it should be variation or combination, then I thought:

¿The triangle formed by the verte $ ABC $ is equal to $ BCA $? Well yes, since that combination of vertex has already been used.

So, now I must know if it is combination with or without repetition

The vertex can be repeated, for example:

$ ABC $ is equivalent to $ CBA $

But,

$ ABC $ is different from $ CBF $, since $ BC $ is repeated but $ F $ is not repeated, so are different triangles.

Ok, I use the formula: $\frac{(n +r - 1)!}{(n - 1)! * r!}$

$n = 6$ ( number of hexagon vertex ), $r = 3$ ( number of triangle vertex)

$= \frac{8!}{5! * 3!} = \frac{336}{!3} = 56$

But the correct is wrong, and the correct result should be $20$.

¿What is my error?

PD:

I used the formula for repetition, and all answer used the other formula, why ?

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It's just $C_3^6=20$. (choosing 3 points from 6).

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  • $\begingroup$ Why without repetition ? Because $ABC$ triangle is different that $ABE$ $\endgroup$ – Eduardo S. Mar 26 '18 at 0:12
  • $\begingroup$ Actually I don’t understand your work. There are only 6 vertices and any combination of three of them form a triangle as no 3 vertices are collinear. What will you get if the hexagon is replaced by a quadrilateral or a pentagon? $\endgroup$ – CY Aries Mar 26 '18 at 1:30
  • $\begingroup$ Exist the formula of combination with repetition and without repetition, why use without repetition, instead of the formula with repetition? $\endgroup$ – Eduardo S. Mar 26 '18 at 1:36
  • $\begingroup$ you use $\frac{n!}{(n-r)! * r!}$ and i am using $\frac{(n+r-1)!}{(n-1)! * r!}$ $\endgroup$ – Eduardo S. Mar 26 '18 at 1:38
  • $\begingroup$ Combination with repetition means that you can use the same object more than once. Your formula includes ‘triangles’ like $AAB$. $\endgroup$ – CY Aries Mar 26 '18 at 1:43
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You should just be selecting three vertices out of six, and ${6 \choose 3}=\frac {6!}{3!(6-3)!}=\frac {720}{6\cdot 6}=20$

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  • $\begingroup$ Can you explain me, why you use $\frac{n!}{(n - r)! * r!}$ formula $\endgroup$ – Eduardo S. Mar 26 '18 at 0:11
  • $\begingroup$ It is just the number of ways to select $r$ things from $n$ without respect to order. You seem to have confused a stars and bars formula with this. $\endgroup$ – Ross Millikan Mar 26 '18 at 0:19
  • $\begingroup$ See my edited, please $\endgroup$ – Eduardo S. Mar 26 '18 at 0:24
  • $\begingroup$ To form a triangle you need three different vertices of the hexagon, so you should not use the formula with repetition. $AAB$ will not form a triangle. $\endgroup$ – Ross Millikan Mar 26 '18 at 3:09

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