How many triangles can you draw using the dots below as vertices?
(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.
Best Answer
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!!