Showing that if six people stand in a ring, then the probability of exactly $t$ people standing between Q and R in the clockwise direction is $1/5$ If they stand in a ring, show that the probability of exactly $t$ people between $Q$ and $R$ in the clockwise direction is $\frac{1}{5}$.
I suppose in this question, we have been given the probability and we need to find the value for $t$. Where $t=$number of people standing between $Q$ and $R$. 
I would like to know that whether this statement 'Clockwise Direction' will have impact on the answer?
Please help me solve this question. 
 A: There are 6! ways to place those people in a circle. Let’s see in how many ways you can place them so that Q is exactly $t, t\in\{0,1,2,3,4\}$ places away from $P$, in clockwise direction. For that:


*

*Place P first, you can do it in 6 ways.

*Place Q next, $t$ places away clockwise. This place is uniquely determined, i.e. there is 1 way of doing that.

*Place the other four people to the remaining 4 places, the number of ways is 4!


So the total number of ways to place those people and get the desired number of places between P and Q, clockwise, is $6\times 4!$.
Thus, assuming all placements are equally likely, the probability is $\frac{6\times 4!}{6!}=\frac{1}{5}$ and it is the same for every $t$.
Now, obviously, if you drop the “clockwise” requirement, then:


*

*$t$ can only be 0,1 or 2

*$t=0$ is now the same as $t=0\lor t=4$ before, so the probabilities add up to $\frac{2}{5}$.

*Similarly, $t=1$ is now the same as $t=1\lor t=3$ before, so the probabilities add up to $\frac{2}{5}$.

*Finally $t=2$ is the same as $t=2$ before (2 people in between, in a circle of 6, means “opposite” - in either direction), so the probability stays $\frac{1}{5}$.

