Question

The number of ways in which 4 students can sit in 7 chairs in a row if there is no empty chair between any two students is
A) 24
B) 28
C) 72
D) 96

Answer 1

Hint:
In the question, we are given a total number of chairs and a total number of students that is the value of \[n\] and \[r\] is given by 7 and 4 respectively. We need to find the number of ways of sitting arrangements such that no chair remains empty between any 2 students which imply that all the 4 students are sitting on conservative chairs. So, the number of ways can be calculated by \[n - r + 1\] and as every student has 4 choices each so \[n - r + 1\] is multiplied by \[4!\].

Complete step by step solution: Consider the data given that is the total number of chairs are given as 7 which means \[n = 7\] and the total number of students are given as 3 which means \[r = 3\].
As we have to calculate the number of ways when no chair remains empty between any 2 students, we have to select 4 conservative chairs between the 4 students.
Thus, we get,
\[
  n - r + 1 = 7 - 4 + 1 \\
   = 4 \\
 \]
Hence, we know that there are 4 students and the number of possibilities of interchanging the 4 chairs would be factorial of 4.
So, the total number of arrangements of the chairs such that no chair remains empty between 4 students are given by \[4 \times 4!\] as there are four students and all 4 students have 24 choices that are \[4!\] choices to made that is why we will multiply 4 by \[4!\].
Thus, we get,
\[
  4 \times 4! = 4 \times \left( {4 \times 3 \times 2 \times 1} \right) \\
   = 96 \\
 \]
Thus, there are 96 ways in which 4 students can sit in a position on 7 chairs such that no seat remains empty between the 2 students.
Hence, option D is correct.

Note: For calculating several ways we use the formula \[n - r + 1\]. In questions like this after calculating the value of \[n - r + 1\], we need to multiply with factorial \[r\], several arrangements can only be final when all the ways are multiplied as in this question, we have 4 students so each student can repeat the process which gives us the total number of arrangements by multiplying 4 by \[4!\].