In how many ways can six study guides be distributed to $15$ students at a circular table so that every student can read the study guide? Fifteen students are sitting around a large circular table for a study session. The teacher has made only six copies of the review guide. No student should get more than one copy of the review guide and any student who does not get one should be able to read a neighbor’s copy. If the students are distinguishable, but the review guides are identical, how many ways are there to distribute the six review guides to the fifteen students subject to these conditions?
First I would find ways to arrange the fifteen students so 
$(15-1)! = 14!$
Then to arrange the $6$ copies of the review guide, I would use combination so 
$15~C~6 = 5005$
But when I multiply these $2$ numbers, I get a huge number.
Do I even need to arrange the fifteen students, or is the answer just $5005$ or is it wrong altogether?
 A: Let the students seat themselves at wish. This can be done in $15!$ ways, resp., in $14!$ ways up to rotations, or in ${1\over2}\cdot14!$ ways up to rotations and reflections.
We now have to distribute the booklets. At the end $6$ students will hold a booklet. In between them there are six slots containing $0$, $1$, or $2$ students each. Denote by $x_i\geq0$ the number of slots containing $i$ students. Then
$$x_0+x_1+x_2=6,\qquad x_1+2x_2=9\ ,$$
hence
$$x_1=3-2x_0,\qquad x_2=3+x_0\ .$$
It follows that $x_0\in\{0,1\}$, so that we obtainthe two admissible solutions
$${\rm (a)}\quad (1,1,4),\qquad{\rm (b)}\quad(0,3,3)\ .$$
(a) As $x_0=1$ there are two adjacent students receiving a booklet. We can choose these in $15$ ways and then the "unary" slot in $5$ ways, makes $75$. 
(b) In order to count the possible arrangements assume that one of the booklets is marked. The marked booklet can be handed out to any of the $15$ students, so that a chain of $14$ students is generated. $5$ among these will get an "ordinary" booklet. The arrangement of slot-sizes between all  booklet holders can be encoded as a word of length $6$ over the alphabet $\{1,2\}$, like so: $221121$. There are ${6\choose3}=20$ such words containing $3$ ones and $3$ twos. Makes ${15\cdot20\over6}=50$ possibilities, the factor $6$ in the denominator coming from the overcounting caused by marking a booklet.
It follows that the booklets can be distributed in $75+50=125$ admissible ways to the $15$ students already sitting around the table..
