Seating Students in a Maths Seminar

Question

I was asked the following question There are six Students $S_1,S_2,S_3,S_4,S_5,S_6$ and on the first day they were given the seats(reserved for them) in the following manner: Student $S_i$ was given seat $R_i$ where $i=1,2...6$.
The next day the organizing committee took an exam in which students were allotted their seats randomly.If the probability that $S_1$ was given the seat $R_1$ and $S_2$ was given $R_6$ and none of the other students(left ones) got the seats preveiously allotted to them(on day 1) is P then find P
for the solution I enumerated all the possilbilies(there were only 11 ways)(and so the ans came out to be ($\frac{11}{6!}$) but I am wondering if there is an alternative to manually writing out all the possibilities.Kindly help me out here.Thankyou!!

Answer 1

$S_1$ occupies $R_1$ and $S_2$ occupies $R_6$. That leaves seats $R_2$ to $R_5$ for students $S_3$ to $S_6$.

Case $1$: If $R_2$ is occupied by $S_6$ then $R_3$ to $R_5$ must be occupied by $S_3$ to $S_5$ and as no student can occupy their previously assigned seats, the number of arrangements is equivalent to the derangement of $3$.

Case $2$: If $R_2$ is occupied by one from $S_3$ to $S_5$ ($3$ ways) then $S_6$ can occupy any of the three remaining seats ($R_3$ to $R_5$) in $3$ ways. The seats of the remaining two students are then fixed. That leads to favorable arrangements as,

$!3 + 3 \cdot 3 = 11$

So, $ ~\displaystyle P = \frac{11}{6!} = \frac{11}{720}$

Answer 2

I think this can work too: Since $S_1$ has got the seat $R_1$ and $S_2$ has got $R_6$ and left students are $S_3$,$S_4$,$S_5$,$S_6$ and left seats to be allotted are $R_2$,$R_3$ ,$R_3$,$R_5$ So I can take 2 cases here 1st of all I can say that all the possibilites come under 2 categories ,one in which $R_2$ is not alloted to $S_6$ and one where $R_2$ is alloted to $S_6$, now for the first part I can "pretend" that $R_2$ is actually R6 so that I can use the method of dearrangemnts (it is used to calculate the no of ways of putting n letters in their corresponding envelops so that no leeter goes in the right envelop) and then I can consider the case when $R_2$ is alloted to $S_6$ and then use dearrangements of 3 things of $R_3$,$R_4$,$R_5$ to $S_3$,$S_4$,$S_5$
this means for the first case no of ways is $$4!(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})$$ for the second case no of ways is $$3!(\frac{1}{2!}-\frac{1}{3!}) $$Adding them up we get 11 cases hence probability is $\frac{11}{6!}$