Counting problem: ways to empty glasses Sorry for my bad english And also my bad math literature in asking my first question.
Assume we have 15 glass full of water.
I need to count the number of possible ways which I can empty 5 glass in a way that no 2 empty glass remain in sequence.
 A: Looking at it in reverse, we can first place $5$ empty glasses in a row. Now we must have $4$ full glasses in between them, to fulfil the constraint. We are now free to distribute remaining $6$ glasses in any of $6$ locations (in between or outside) the empty glasses. 
So this is equivalent to putting $6$ identical balls in $6$ different boxes, where some boxes can even be empty...  Which is $C(6+6-1, 6) = C(11, 6)$.
A: If the glasses are in a line, say we draw a full glass as a star $*$, and an empty glass as a vertical bar $|$. Then a possible way to empty five glasses is drawn as a line of five stars and ten bars, where the bars are not next to each other. One such way could be drawn as
$|**|**|**|**|**$
This is similar to the problem discussed as Theorem One in Stars and bars notation, except the bars can be placed on the ends as well. We can solve the problem by considering four cases:


*

*There is no bar on either the left end or the right end. This is then the same as Theorem One in the link, with $k=10$ and $n=6$, so there are $\binom{9}{5}$ of these.

*There is a bar on the left end. Then ignore this bar, and the rest of the line follows Theorem One with $k=10$ and $n=5$, so there are $\binom{9}{4}$ of these.

*There is a bar on the right end. Same as case 2, $\binom{9}{4}$ of these.

*There is a bar on both ends. Ignore them both, and now $n=4$, so $\binom{9}{3}$ of these.
The total number of arrangements is then $\binom{9}{5} + 2\binom{9}{4} + \binom{9}{3} = \binom{10}{5} + \binom{10}{4} = \binom{11}{5}.$
