Question

Let $C_1,C_2,...C_n,...$ be a sequence of concentric circles. The nth circle has the radius n and it has n openings. A point P starts travelling on the smallest circle $C_1$ and leaves it at an opening along the normal at the point of opening to reach the next circle $C_2$. Then it moves on the second circle $C_2$ and leaves it likewise to reach the third circle $C_3$ and so on. The total number of different paths in which the point can come out of the nth circle is

• A. $2^n\cdot n!$
• B. $2^{n-1}\cdot n!$
• C. $n!$
• D. $2^{n-1}\cdot (n-1)!$

Answer 1

The correct option is A $2^n\cdot n!$

Given, it moves along a circle
So,for a circle with n openings it will have n choices to select one of these openings.But, to leave an opening, it can move clockwise or counter clockwise
It makes '2n' ways for leaving a circle
$\therefore$ The total number of different paths in which the point can come out of the nth circle $=2\left(4\right)\left(6\right)........\left(2n\right)$
$=2^n\times n!$