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)

$3\cdot {}^n{P}_4={}^{n-1}{P}_5\Rightarrow 3\cdot \frac{n!}{(n-4)!}=\frac{(n-1)!}{(n-6)!}$
$3n=(n-4)(n-5)\Rightarrow 3n=n^2-9n+20\Rightarrow n^2-12n+20=0$
$\Rightarrow (n-10)(n-2)=0\Rightarrow n=10\text{ as }n\ne 2$
Any 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 $=10\cdot {}^5{C}_1=50$