Question

A train is having $12$ stations on route. It has to be stopped at $4$ stations. Find the number of ways it can be stopped if no two stopping stations are consecutive.
A. ${}^{8}{{C}_{4}}$
B. ${}^{9}{{C}_{4}}$
C. ${}^{12}{{C}_{4}}-{}^{8}{{C}_{4}}+{}^{4}{{C}_{4}}$
D. ${}^{12}{{C}_{4}}-{}^{12}{{C}_{4}}+{}^{8}{{C}_{4}}-{}^{6}{{C}_{4}}$

Answer 1

Hint: To find the number of ways the train can be stopped at $4$ stations if no two stopping stations are consecutive, we will find the non-stopping stations. Number of non-stopping stations $=12-4=8$ . The 4 stations can be arranged within these 8 stations such that no two stopping stations are consecutive or adjacent. Let the non-stopping stations be A, B, C, D, E, F, G, H. So the stopping stations can be arranged as \[\bullet A\bullet B\bullet C\bullet D\bullet E\bullet F\bullet G\bullet H\] where each dot is the positions where the 4 stations can be placed, that is, 9. Using combination, we can find the number of ways in which the 4 stations can be arranged.

Complete step by step answer:
We need to find the number of ways the train can be stopped at $4$ stations if no two stopping stations are consecutive.
It is given that the total number of stations is $12$ . Of these, the train has to be stopped at $4$ stations. So, let us consider the remaining stations, that is, $12-4=8$ stations.
The 4 stations can be arranged within these 8 stations in the following manner such that no two stopping stations are consecutive or adjacent.

In the figure, the non-stopping stations are denoted as A, B, C, D, E, F, G, H. The \[\bullet \] sign denotes the positions where the stopping stations can lie.
From the figure, there are 9 such positions. So from these 9 positions, we have to place only 4 stations.
To find the number of ways this can be done, we will use the combination.
So, out of the 9 positions, 4 stations can be located in ${}^{9}{{C}_{4}}$ ways.

So, the correct answer is “Option B”.

Note:
To find the number of ways in which something can be arranged when the order doesn’t matter, we use combination not permutation. When the order is a concern, we go for permutation. Do not get confused with the representations used. The combination is denoted as ${}^{n}{{C}_{r}}$ while the permutation is represented as ${}^{n}{{P}_{r}}$.