
There are p intermediate stations on a railway line. In how many ways can a train stop at 3 of these stations if no 2 stations are consecutive?
Answer
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Hint: To solve this question, we should write a linear equation in terms of x, y, z, t of the below diagram. Let us consider the two stations to be A and B. Let us consider that the train stops at the stations 1 and 2 and 3. The variables x, y, z, t are the number of stations which the train crosses between A and station-1, station-1 and station-2, station-2 and station-3, station-3 and B respectively. We can infer that excluding the three stopping stations, there will be $p-3$ stations which train passes without stopping. We can write that $x+y+z+t=p-3$. As the stations should not be consecutive, we can infer that the minimum values of y and z should be 1. The values of x and t can be zero or greater than zero as the train can immediately stop after it crosses A and it can even stop at the just before station of B. So, we can write from the equation that
$x+\left( y-1 \right)+\left( z-1 \right)+t=p-3-2=p-5$. By naming $y-1$and $x-1$ as another two variables, we get an equation in which there are four variables and zeros are allowed in those variables. The number of solutions of the equation
$x+y+z+p+....r\text{ variables}=n$ where zeros are allowed is given by ${}^{n+r-1}{{C}_{r-1}}$. Using this formula, we get the answer.
Complete step-by-step answer:
Let us consider the two stations to be A and B. Let us consider that the train stops at the stations 1 and 2 and 3. The variables x, y, z, t are the number of stations which the train crosses between A and station-1, station-1 and station-2, station-2 and station-3, station-3 and B respectively. We can infer that excluding the three stopping stations, there will be $p-3$ stations which train passes without stopping. We can write the total number of stations in which the train doesn’t stop from the figure as
$x+y+z+t=p-3\to \left( 1 \right)$.
The value of x can be anything starting from zero as the train can stop in the immediate station of A. SO, we can write that
$x\ge 0$
We can infer from the question that the stations should not be consecutive, which means that there should be at least one station between the stations at which the train stops. From the definitions of the variables, we can infer that the number of intermediate stations between the stations at which the train stops are y and z. So, we can write that
$y\ge 1,z\ge 1$
We can write the above inequalities as
$y-1\ge 0,z-1\ge 0$.
The value of t can be anything as the train can even stop at the station just before the station B. So,
$t\ge 0$
We can write the equation-1 as
$x+\left( y-1 \right)+\left( z-1 \right)+t=p-3-2=p-5$
Let us define new variables which are
$\begin{align}
& y-1={{y}_{1}}\ge 0 \\
& z-1={{z}_{1}}\ge 0 \\
\end{align}$
The above equation becomes
$x+{{y}_{1}}+{{z}_{1}}+t=p-5$
We can infer that each of the variables of the above equation can be any number greater than or equal to zero.
The number of solutions of the equation
$x+y+z+p+....r\text{ variables}=n$ where zeros are allowed is given by ${}^{n+r-1}{{C}_{r-1}}$.
Using this formula where $n=p-5$ and $r=4$, we get
Number of ways = ${}^{p-5+4-1}{{C}_{4-1}}={}^{p-2}{{C}_{3}}$
$\therefore $ The number of ways for the train to stop at three non-consecutive stations is ${}^{p-2}{{C}_{3}}$
Note: We can develop a general formula for this kind of question. When the number of stoppages is 3, we get a value of r as $r=4$. So, when the number of stoppages are x, we get $r=x+1$. We can write the equation as
$x+y+z+a+b...+t=p-x$. We can infer that there are two values which are greater than or equal to 1 when the stoppages are 3. So, there will be $x-1$ variables, which have their value greater than or equal to 1. We can write the equation as
$x+{{y}_{1}}+{{z}_{1}}+{{a}_{1}}+{{b}_{1}}...+t\left( x+1\text{ }terms \right)=p-x-\left( x-1 \right)=p+1-2x$. The number of solutions will be ${}^{p+1-2x+x+1-1}{{C}_{x+1-1}}={}^{p+1-x}{{C}_{x}}$. So, we can write that the number of ways in which a train can stop at x non-consecutive stops is ${}^{p+1-x}{{C}_{x}}$ .
$x+\left( y-1 \right)+\left( z-1 \right)+t=p-3-2=p-5$. By naming $y-1$and $x-1$ as another two variables, we get an equation in which there are four variables and zeros are allowed in those variables. The number of solutions of the equation
$x+y+z+p+....r\text{ variables}=n$ where zeros are allowed is given by ${}^{n+r-1}{{C}_{r-1}}$. Using this formula, we get the answer.
Complete step-by-step answer:

Let us consider the two stations to be A and B. Let us consider that the train stops at the stations 1 and 2 and 3. The variables x, y, z, t are the number of stations which the train crosses between A and station-1, station-1 and station-2, station-2 and station-3, station-3 and B respectively. We can infer that excluding the three stopping stations, there will be $p-3$ stations which train passes without stopping. We can write the total number of stations in which the train doesn’t stop from the figure as
$x+y+z+t=p-3\to \left( 1 \right)$.
The value of x can be anything starting from zero as the train can stop in the immediate station of A. SO, we can write that
$x\ge 0$
We can infer from the question that the stations should not be consecutive, which means that there should be at least one station between the stations at which the train stops. From the definitions of the variables, we can infer that the number of intermediate stations between the stations at which the train stops are y and z. So, we can write that
$y\ge 1,z\ge 1$
We can write the above inequalities as
$y-1\ge 0,z-1\ge 0$.
The value of t can be anything as the train can even stop at the station just before the station B. So,
$t\ge 0$
We can write the equation-1 as
$x+\left( y-1 \right)+\left( z-1 \right)+t=p-3-2=p-5$
Let us define new variables which are
$\begin{align}
& y-1={{y}_{1}}\ge 0 \\
& z-1={{z}_{1}}\ge 0 \\
\end{align}$
The above equation becomes
$x+{{y}_{1}}+{{z}_{1}}+t=p-5$
We can infer that each of the variables of the above equation can be any number greater than or equal to zero.
The number of solutions of the equation
$x+y+z+p+....r\text{ variables}=n$ where zeros are allowed is given by ${}^{n+r-1}{{C}_{r-1}}$.
Using this formula where $n=p-5$ and $r=4$, we get
Number of ways = ${}^{p-5+4-1}{{C}_{4-1}}={}^{p-2}{{C}_{3}}$
$\therefore $ The number of ways for the train to stop at three non-consecutive stations is ${}^{p-2}{{C}_{3}}$
Note: We can develop a general formula for this kind of question. When the number of stoppages is 3, we get a value of r as $r=4$. So, when the number of stoppages are x, we get $r=x+1$. We can write the equation as
$x+y+z+a+b...+t=p-x$. We can infer that there are two values which are greater than or equal to 1 when the stoppages are 3. So, there will be $x-1$ variables, which have their value greater than or equal to 1. We can write the equation as
$x+{{y}_{1}}+{{z}_{1}}+{{a}_{1}}+{{b}_{1}}...+t\left( x+1\text{ }terms \right)=p-x-\left( x-1 \right)=p+1-2x$. The number of solutions will be ${}^{p+1-2x+x+1-1}{{C}_{x+1-1}}={}^{p+1-x}{{C}_{x}}$. So, we can write that the number of ways in which a train can stop at x non-consecutive stops is ${}^{p+1-x}{{C}_{x}}$ .
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