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Q. Let $A_i$ where $i=1, \ldots \ldots \ldots \ldots . .12$ are the vertices of a regular dodecagon and $G$ is its centre. Let ' $s$ ' be the number of straight lines that can be formed with these 13 points, ' $t$ ' be number of triangle that can be formed and ' $d$ ' be number of diagonal in a dodecagon. Find the value of $(s+t+d)$.

Permutations and Combinations

Solution:

$s={ }^{12} C _2=66$
$t ={ }^{13} C _3-6=\frac{13 \cdot 12 \cdot 11}{6}-6=286-6=280 $
$d ={ }^{12} C _2-12=66-12=54$
Hence, $( s + t + d )=66+280+54=120+280=400$.