Question 27

My logic was that after every step king has 8 options and that the total number of different paths would be equal to the number of ways he could come back

Question 27

My logic was that after every step king has 8 options and that the total number of different paths would be equal to the number of ways he could come back

\bullet\; One Direction:

\displaystyle\bigg(3\text{F},3\text{B}\bigg)=\binom{4}{1}\cdot \frac{6!}{3!\cdot 3!}=80.

\bullet\; Two Direction:

\displaystyle\bigg(2\text{F},2\text{B}\bigg)+\bigg(1\text{R},1\text{L}\bigg)=\frac{6!}{2!\cdot 2!}\cdot \binom{4}{2}\cdot 2=2160.

\bullet\; Three Direction:

\displaystyle\bigg(1\text{F},1\text{B}\bigg)+\bigg(1\text{R},1\text{L}\bigg)+\bigg(1\text{D},1\text{DBa}\bigg)=6!\cdot \binom{4}{3}=2880.

So we get total ways

\color{brown}=80+2160+2880=5120.

Note: Solution is not completely mine. I have Taken some help from my friend.

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In between Brackets, There is no plus sign.

Done by mistake.

Diagram for 2 nd and 3 rd cases.

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@Jagdish_Singh. Sir In second and third steps what are the 4C2 and 4C3 respectively for ?

I understand that 6!/(2!*2! ) Is essentially to arrange the steps