Can someone help me on this permutations and combination question?

"The diagram shows 11 points marked in two rows". We can't see the diagram! How many points are in each row?

Suppose there are "n" points in the first row and 11- n points in the second row. A can be any one of the n points and b can be any one of the other n- 1 points. That would make n(n-1) possible A,B pairs except that swapping A and B would give the same triangle so there are actually n(n-1)/2 pairs, A triangle would be A and B connected to any of the 11- n points on the second line. That is n(n-1)(11-n)/2 triangles. You say the answer is 7. Is there an integer n so that n(n-1)(11- n)/2= 7? That is the same as n(n-1)(11- n)= 2*7. Both 2 and 7 are prime so we must have one of the three factors 1, another 2, and another 7. Clearly we would have to have n= 2 so that n-1= 1 but then 11- n= 9, not 7! There is no way you can have "11 points marked in two rows" and have 7 triangles.

Did you mean that the two rows have 11 points each? In that case, using the same argument as above There are 11(11-1)/5= 22 possibilities for AB then 22(11)= 242 triangles.

Please tell us how many points there are in each row!
 
Dr H,
I am a bit confused by your work. If A and B are marked on the 1st row, then why are you computing nC2? I would think that there is only 1 way to choose A and B. What am I missing?

I would think that for there to be 7 triangles with two vertices being A and B is if the 2nd row has 7 points and the 1st row has 4 points, A, B, C and D.
Hmm, now I think that you are not correct(??)
 
HallsofIvy said:
"The diagram shows 11 points marked in two rows". We can't see the diagram! How many points are in each row?

Suppose there are "n" points in the first row and 11- n points in the second row. A can be any one of the n points and b can be any one of the other n- 1 points. That would make n(n-1) possible A,B pairs except that swapping A and B would give the same triangle so there are actually n(n-1)/2 pairs, A triangle would be A and B connected to any of the 11- n points on the second line. That is n(n-1)(11-n)/2 triangles. You say the answer is 7. Is there an integer n so that n(n-1)(11- n)/2= 7? That is the same as n(n-1)(11- n)= 2*7. Both 2 and 7 are prime so we must have one of the three factors 1, another 2, and another 7. Clearly we would have to have n= 2 so that n-1= 1 but then 11- n= 9, not 7! There is no way you can have "11 points marked in two rows" and have 7 triangles.

Did you mean that the two rows have 11 points each? In that case, using the same argument as above There are 11(11-1)/5= 22 possibilities for AB then 22(11)= 242 triangles.

Please tell us how many points there are in each row!
Click to expand...
I've solved it

We know that there are 11 point. 4 point on top, 7point bottom. On the top there are A B C and D.

For a ) point A and B are the two vertices of the triangle. So we already have two vertices and we need one more. The equation is 7C1 because it can be any one point on the bottom.

For b ) Point C is one of the vertices. So we need two more point which can be 1 on top and 1 on the bottom / or 2 at the bottom to make a triangle. The equation is 3C1*7C1 + 7C2
 
Jomo said:
Dr H,
I am a bit confused by your work. If A and B are marked on the 1st row, then why are you computing nC2? I would think that there is only 1 way to choose A and B. What am I missing?

I would think that for there to be 7 triangles with two vertices being A and B is if the 2nd row has 7 points and the 1st row has 4 points, A, B, C and D.
Hmm, now I think that you are not correct(??)
Click to expand...
For a ) point A and B are the two vertices of the triangle. So we already have two vertices and we need one more. The equation is 7C1 because it can be any one point on the bottom.

For b ) Point C is one of the vertices. So we need two more point which can be 1 on top and 1 on the bottom / or 2 at the bottom to make a triangle. The equation is 3C1*7C1 + 7C2
 
Please do not say that points A and B are the two vertices of the triangle. There are three vertices of a triangle. Rather say points A and B are two vertices of a triangle.
 
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