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 made using the vertices of the polygon of sides ' $n$ ' if exactly two adjacent sides of the quadrilateral are common to the sides of the $n$-gon, is

• A. 50
• B. 60
• C. 70
• D. None

Answer 1

Answer: (A)

$3 \cdot{ }^n P_4={ }^{n-1} P_5 \Rightarrow 3 \cdot \frac{n !}{(n-4) !}=\frac{(n-1) !}{(n-6) !}$
$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 3 consecutive vertices can be selected in 10 ways (any two adjacent sides)
$(1,2,3) \text { or }(2,3,4) \text { or }$
or $(10,1,2)$
say e.g. $(1,2,3)$
now 4 and 10 cannot be selected
from the remaining five vertices from 5 to 9 any one vertex can be selected in ${ }^5 C _1$ ways.
$\therefore $ number of such quadraliterals $=10 \cdot{ }^5 C _1=50$