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)

$3 \cdot{ }^n P_4={ }^{n-1} P_5 \Rightarrow 3 \cdot \frac{n !}{(n-4) !}=\frac{(n-1) !}{(n-6) !}$
$3 n =( n -4)( n -5) \Rightarrow 3 n = n ^2-9 n +20 \Rightarrow n ^2-12 n +20=0 $
$\Rightarrow ( n -10)( n -2)=0 \Rightarrow n =10 \text { as } n \neq 2$
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 $=10 \cdot{ }^5 C _2=100$