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How many triangles can you draw using the dots below as vertices?

enter image description here

(a) Find an expression for the answer which is the sum of three terms involving binomial coefficient.

(b) Find an expression for the answer which is the difference of two binomial coefficient.

(c) Generalize the above to state and prove a binomial identity using a combinatorial proof. Say you have $x$ points on the horizontal axis and $y$ points in the semi-circle.

Please can someone help me in these kind of sums!!

My work

Finally I got it first from five semicircle points select any 3 points to be vertices of triangle =10 ways second from 7 horizontal points select any 2 points and from 5 semicircle points select any 1 point = 21*5 = 105 ways Third from 7 horizontal points select any 1 points and from 5 semicircle points select any 2 point = 70 ways Thus 70 + 105 + 10=185 triangles are possible.

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  • $\begingroup$ Is there supposed to be a picture here? $\endgroup$ – saulspatz Aug 16 at 16:20
  • $\begingroup$ @RobertZ pings don't work in this way. Generally that's an unproductive way to approach such matters. (I removed the comment, you still get the ping as you edited the question.) $\endgroup$ – quid Aug 16 at 20:43
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Hint for (a). Any proper triangle will have $0$, $1$, or $2$ points along the horizontal line.

Hint for (b). A proper triangle will have the $3$ vertices not all along the same line. Now we have $12$ points of which $7$ are along the horizontal line.

Now it's your turn!!

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    $\begingroup$ I'm not sure I understand your hint for (b). I would say: which sets of 3 dots do not make a triangle? $\endgroup$ – Henning Makholm Aug 16 at 16:55
  • $\begingroup$ x @Robert: Unfortunately I'm still confused. Was it meant to be a hint for (c) instead? That's the question that introduces meanings for $x$ and $y$? $\endgroup$ – Henning Makholm Aug 16 at 17:00
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    $\begingroup$ Any hint for a smart way to arrive at answers to (a) and (b) will automatically be a hint for the general case (c) as well ... $\endgroup$ – Hagen von Eitzen Aug 16 at 17:11
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finally I got it first from five semicircle points select any 3 points to be vertices of triangle =10 ways second from 7 horizontal points select any 2 points and from 5 semicircle points select any 1 point = 21*5 = 105 ways Third from 7 horizontal points select any 1 points and from 5 semicircle points select any 2 point = 70 ways Thus 70 + 105 + 10=185 triangles are possible.

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  • $\begingroup$ OK. This is (a). BTW you are supposed to write your progress under your question and not as an answer. Otherwise your question will be closed! Now try with (b) and (c). $\endgroup$ – Robert Z Aug 16 at 18:35

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