Question

There are $p$ intermediate stations on a railway line from one terminus to another. Let ${}^{(p-k)}{C}_m$ be the number of ways a train can stop at $3$ of these intermediate stations if no two of these stopping stations are to be consecutive.Find $k+m$?

Answer 1

Let there be $p$ intermediate stations between two terminus stations $A$ and $B$ as shown in the figure.
No. of ways the train can stop in three intermediate stations $={}^p{C}_3$
These are comprised of two consecutive cases viz.
$\left(i\right)$ At least two stations are consecutive
$\left(ii\right)$ No two of which is consecutive
Now there are $(p-1)$ pairs of consecutive intermediate stations.
In order to get a station trio in which at least two stations are consecutive, each pair can be associated with a third station in $(p-2)$ ways. Hence total no. of ways in which $3$ stations consisting of at least two consecutive stations can be chosen in $(p-1)(p-2)$ ways. Among these, each triplet of consecutive stations occur twice. For example, the pair $(S_n,S_{n-1})$ when combined with $S_{n+1}$ and the pair $(S_n,S_{n+1})$ when combined with $S_{n-1}$ gives the same triplet and is counted twice. So, the number of three consecutive stations trio should be subtracted.
Now, no. of these three consecutive stations trio is $(p-2)$.
Hence, the number of ways the train can stop in three consecutive stations is
$={}^p{C}_3-(p-2{)}^2$
$=\frac{p(p-1)(p-2)}{\mathrm{1.2.3}}-(p-2{)}^2$
$=(p-2)\left[\frac{p^2-p-6p+12}{6}\right]$
$=\frac{(p-2)(p^2-7p+12)}{6}$
$=\frac{(p-2)(p-3)(p-4)}{\mathrm{1.2.3}}$
$={}^{(p-2)}{C}_3$
$k+m=2+3=5$