Question

Consider a polygon of sides ' $n$ ' which satisfies the equation, $3\cdot {}^n{P}_4={}^{n-1}{P}_5$
Number of quadrilaterals that can be formed using the vertices of a polygon of sides ' $n$ ' if exactly 1 side of the quadrilateral is common with the side of the $n$-gon, is

• A. 150
• B. 100
• C. 96
• D. None

Answer 1

Answer: (B)

$3\cdot {}^n{P}_4={}^{n-1}{P}_5\Rightarrow 3\cdot \frac{n!}{(n-4)!}=\frac{(n-1)!}{(n-6)!}$
$3n=(n-4)(n-5)\Rightarrow 3n=n^2-9n+20\Rightarrow n^2-12n+20=0$
$\Rightarrow (n-10)(n-2)=0\Rightarrow n=10\text{ as }n\ne 2$
Any 2 consecutive vertices can be selected in 10 ways
$(1,2)$ or $(2,3)$ or or $(10,1)$
say e.g. $(1,2)$
now 3 and 10 cannot be selected and from the remaining 6 vertices any two non-consecutive vertices can be slected in ${}^5{C}_2$ ways.
$\therefore$ number of such quadraliterals $=10\cdot {}^5{C}_2=100$