(a)
Hexagon has six vertices.
To form a triangle, any three vertices are required.
So, the number of ways is
\(\begin{aligned} {}^nC_{r} &= \dfrac{6!}{3!(6-3)!} \\\\ &=\dfrac{6!}{3!3!}\\\\ &=20. \end{aligned}\)
(b)
Number of ways to choose two organic motifs and one geometric motif, \({}^4C_{2} \times{}^5C_{1}\).
Number of ways to choose one organic motif and two geometric motifs, \({}^4C_{1} \times {}^5C_{2}\).
So, the number of ways
\(({}^4C_{2} \times {}^5C_{1}) + ({}^4C_{1} \times {}^5C_{2}) = 70\).
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