Finding the number of consecutive objects Thee are 14 intermediate stations between A and B. A train is to be scheduled from point B to A so that it halts at exactly three intermediate stations, no 2 of which can be consecutive. The number of ways this can be accomplished is ?
My solution involved ${14 \choose 3}$ - 13 ( as thirteen was the number of consecutive pairing of two numbers possible.
Where am I going wrong ?
 A: Let the number of stations between the starting point and the first halt be $x_1$, between the first and second halt be $x_2$, between the second and third halt be $x_3$, and between the third halt and the destination be $x_4$.
Now,
$$x_1+x_2+x_3+x_4+3\text{(the three stations where the train halted)}=14$$
That is,
$$x_1+x_2+x_3+x_4=11$$
Now, no two of these stations will be consecutive if $x_i\ge1$ where $i\in\{2,3\}$
Let $x_2=y_2+1$ and $x_3=y_3+1$. Then, we have $y_2,y_3\ge0$.
Let $x_1=y_1,x_2=y_2$
Thus, the solution will be the number of non negative integer solutions of the equation,
$$y_1+y_2+y_3+y_4=9$$
A: Method 1:  We use eleven blue balls to represent the bypassed stations and three red balls to represent the intermediate stations where the train stops. Line up the eleven blue balls in a row, leaving gaps between them in which to insert the red balls.  There are twelve spaces for the red balls, in the ten spaces between successive blue balls and the two spaces at the ends of the row.  Select three of these twelve spaces in which to place the red balls.  Now label the balls from left to right with the names of the intermediate stations in the order the train encounters them.  The names on the red balls are the three stations where the train stops.  Since there are $\binom{12}{3}$ ways of selecting the positions of the red balls, there are also $\binom{12}{3}$ ways of selecting three intermediate stops for the train so that no two of them are consecutive.
Method 2:  We fix your attempt by using the Inclusion-Exclusion Principle.
There are $\binom{14}{3}$ ways of selecting three of the $14$ intermediate stops.  However, we must exclude those in which at least two of the stops are consecutive.
There are $13$ pairs of two consecutive intermediate stops (since the first of the two consecutive stops cannot be the last stop before the final destination).  For each such choice, there are $12$ ways of choosing the third intermediate stop (which you forgot to take into account).  Hence, there are $13 \cdot 12$ ways to select ctwo consecutive stations.
However, if we subtract $13 \cdot 12$ from $\binom{14}{3}$, we will have subtracted each selection of three consecutive intermediate stops twice, once when we select the first two of the three consecutive intermediate stops and once when we choose the last two of the three consecutive intermediate stops.  Since we only want to exclude these selections once, we must add the number of ways in which three consecutive stops can be selected.  There are $12$ ways of selecting three consecutive intermediate stops (since the first of the three consecutive stops cannot be on of the last two stops before the final destination).  
By the Inclusion-Exclusion Principle, the number of ways three of the fourteen intermediate stops can be selected so that no two of them are consecutive is 
$$\binom{14}{3} - 13 \cdot 12 + 12 = \binom{12}{3}$$  
A: Another way:
With "non-stopping" station as $\color{green}{\Large\bullet}$ and a "stopping" station as $\color{red}{\Large \bullet}$, add an extra $\color{green}{\Large\bullet}$ 
Create $3$ boxes, $\boxed{\color{green}{\Large\bullet}\color{red}{\Large\bullet}}\quad\boxed{\color{green}{\Large\bullet}\color{red}{\Large\bullet}}\quad\boxed{\color{green}{\Large\bullet}\color{red}{\Large\bullet}}\,\;$now there are $12$ objects in total, 
Place the boxes in $\binom{12}{3}$ ways, and remove the extra $\color{green}{\Large\bullet}$ from the first box.
