Question

Three-person enter a railway compartment having 5 vacant seats. In how many ways can they seat themselves?

Answer 1

Hint:For solving this question, we will directly use the formula of the number of ways to select $r$ objects from the $n$ distinct objects with values of $n=5$ and $r=3$ . After that, we will apply the formula for the total number of ways of arranging $r$ distinct objects in a linear arrangement with values of $r=3$ . Then, we will multiply both the result and solve for the number of different ways in which they can seat themselves.

Complete step-by-step answer:
Given:
Three-person enter a railway compartment having 5 vacant seats. And we have to find the total number of ways in which they can seat themselves.
Now, before we proceed we should know the following important concept and formulas:
1. Fundamental Principle of Multiplication: If there are two jobs such that one of them can be completed in $m$ ways, and when it has been completed in any of these $m$ ways, the second job can be completed in $n$ ways. Then, two jobs in succession can be completed in $m\times n$ ways.
2. Number of ways to select $r$ objects from the $n$ distinct objects. The formula for the number of different possible ways of selection is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
3. Number of linear arrangements of $r$ distinct objects will be equal to $r!$ .
Now, we will accomplish the given task by doing two jobs in succession. The jobs are mentioned below:
First Job:
Now, first, we will select $3$ seats for $3$ persons form the $5$ vacant seats, with the help of the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . Then, the number of ways in which we can select $3$ seats from $5$ seats will be $m={}^{5}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!}=\dfrac{1\times 2\times 3\times 4\times 5}{1\times 2\times 3\times \left( 2! \right)}=\dfrac{20}{1\times 2}=10$ ways.
Second Job:
Now, as the persons are seated, we will find the number of ways of arranging $3$ persons among themselves, with the help of the formula $r!$ . Then, the number of ways in which $3$ persons can be arranged among themselves will be $n=3!=1\times 2\times 3=6$ ways.
Now, we can say that, if we have to find the number of ways in which $3$ persons can be seated on $5$ seats. First, we will select three seats for the persons in $m$ ways, then we will arrange the 3 persons among themselves in $n$ ways. And from the fundamental principle of multiplication, we can say that the total number of ways possible will be equal to $m\times n$ ways. Then,
Total number of ways will be $=m\times n=10\times 6=60$ ways.
Thus, the required number of ways is 60.

Note: Here, the student before solving first try to understand the problem and divide the given task into two jobs then, we should apply the concept of the fundamental principle of multiplications and suitable formulas correctly and calculate the correct answer. Moreover, we could have solved this problem easily by the following method:
As there are 3 persons and 5 seats. So, we can see the number of seats available for every person if they sit subsequently. Then, for the first person, there will be $5$ seats, for the second person there will be $4$ seats and for the third person, there will be $3$ seats. Then, the total number of ways will be $5\times 4\times 3=60$ ways.Also we can calculate directly by applying the the formula for the number of different possible ways i.e ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ .