Question:
Find the number of
(i) diagonals (ii) triangles formed in a decagon.
Solution:
A decagon has 10 sides.
(i) Number of diagonals $=\frac{n(n-3)}{2}=\frac{10(10-3)}{2}=35$
(ii) Number of triangles (i.e. 3 sides are to be selected) $={ }^{10} C_{3}=\frac{10}{3} \times \frac{9}{2} \times \frac{8}{1}=120$
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