There are 12 intermediate stations between two places A and B. Find the number of ways in which a train can be made to stop at 4 of these intermediate stations so that no two stopping stations are consecutive.
Answer
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Hint: In this question we need to find the number of ways a train travels between two stations, given some constraints. In order to solve this question, we will assume the number of stations missed before every stopping station as ${x_i}$. This will help us simplify the question and reach the answer.
Complete step-by-step answer:
We have been given that there are 12 intermediate stations between two places A and B and trains will stop only at four stations which are not consecutive.
So, let ${x_1}$ be the number of stations before Stop 1, ${x_2}$ be the number of stations before Stop 2, ${x_3}$ be the number of stations before Stop 3 and ${x_4}$ be the number of stations before Stop 4. Also, ${x_5}$ be the number of stations on the right of 5th station.
So, ${x_1} \geqslant 0,{x_2} \geqslant 1,{x_3} \geqslant 1,{x_4} \geqslant 1$
And $ \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} + {x_5} = 8$
So, let ${x_2} - 1 = {y_2},{x_2} - 1 = {y_3},{x_4} - 1 = {y_4}$
$ \Rightarrow {x_1} + {y_2} - 1 + {y_3} - 1 + {y_4} - 1 + {x_5} = 8$
$ \Rightarrow {x_1} + {y_2} + {y_3} + {y_4} + {x_5} = 5$
So, the number of solutions of this equation using the formula; ${\text{Number of ways}} = {}^{n + r - 1}{C_{r - 1}}$,
where n = 5, r = 5
$ \Rightarrow {}^{5 + 5 - 1}{C_{5 - 1}} = {}^9{C_4}$
Now, as ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
So, ${}^9{C_4} = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}} = \dfrac{{9!}}{{4!5!}} = 126$
Hence, there are 126 ways in which a train can be made to stop at 4 of these intermediate stations so that no two stopping stations are consecutive.
Note: Whenever we face such types of problems the value point to remember is that we need to have a good grasp over combinations and its formulas. The formula to calculate the required combination has been discussed above. This helps in getting us the required expressions and gets us on the right track to reach the answer.
Complete step-by-step answer:
We have been given that there are 12 intermediate stations between two places A and B and trains will stop only at four stations which are not consecutive.
So, let ${x_1}$ be the number of stations before Stop 1, ${x_2}$ be the number of stations before Stop 2, ${x_3}$ be the number of stations before Stop 3 and ${x_4}$ be the number of stations before Stop 4. Also, ${x_5}$ be the number of stations on the right of 5th station.
So, ${x_1} \geqslant 0,{x_2} \geqslant 1,{x_3} \geqslant 1,{x_4} \geqslant 1$
And $ \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} + {x_5} = 8$
So, let ${x_2} - 1 = {y_2},{x_2} - 1 = {y_3},{x_4} - 1 = {y_4}$
$ \Rightarrow {x_1} + {y_2} - 1 + {y_3} - 1 + {y_4} - 1 + {x_5} = 8$
$ \Rightarrow {x_1} + {y_2} + {y_3} + {y_4} + {x_5} = 5$
So, the number of solutions of this equation using the formula; ${\text{Number of ways}} = {}^{n + r - 1}{C_{r - 1}}$,
where n = 5, r = 5
$ \Rightarrow {}^{5 + 5 - 1}{C_{5 - 1}} = {}^9{C_4}$
Now, as ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
So, ${}^9{C_4} = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}} = \dfrac{{9!}}{{4!5!}} = 126$
Hence, there are 126 ways in which a train can be made to stop at 4 of these intermediate stations so that no two stopping stations are consecutive.
Note: Whenever we face such types of problems the value point to remember is that we need to have a good grasp over combinations and its formulas. The formula to calculate the required combination has been discussed above. This helps in getting us the required expressions and gets us on the right track to reach the answer.
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