Quote:
Originally Posted by Durba
This is a common Q, however i am stuck at the approach. Why 1 method works and the other doesnt.
Q > Find the number of triangles whose vertices are vertices of an octagon but none of the sides of such triangle are taken from the sides of the octagon.
My approach >
the Q can be visualized as the number of ways of choosing vertices so that no two are consecutive. If the vertices are a,b,c,d,e,f,g,h and i pick "a" then let there be X1 vertices between "a" and the next vertices. Similarly X2 be the number of vertices b/w 2nd & 3rd vertices and X3 be number of vertices between 3rd and first vertex. In that case, X1+X2+X3 = 5 -------------- (1)
the number of ways vertices can be chosen will be equal to the number of solution of the eq1 sbove. where x1,x2,x3 > 0
Let x1 = y1 + 1, X2 = Y2 + 1, X3 = Y3 + 1.
then eq 1 changes to Y1+Y2+Y3 = 2, (y1,Y2,y3>=0)
And the number of ways of solving eq2 (by the book) is (2+3-1)C(3-1) = 4C2 = 6.
Also the number of vertices can be chosen in 8 ways ...
so total num of ways = 6*8 = 48.
But the answer given in the book is 16 & their approach is also different.
Any feedback on where my approach is wrong ?
If you want i can also post the solution given in the book but please also hav a look at my approach above and please tell me where i am mistaken. Thank you.
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I think the approach should be
choose 3 pints from 8 points = 8c3 = 56
choose 3 points such 2 sides are side of the octagon = 8
choose 3 points such that 1 side is side of octagon = 4c1 * 8 = 32
so total 56 - 32 -8 = 40
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