Question

In an examination hall, there are four rows of chairs. Each row has 8 chairs one behind the other. There are two classes sitting for the examination with 16 students in each class. It is desired that in each row, all students belong to the same class and that no two adjacent rows are allotted to the same class. In how many ways can these 32 students be seated?

Answer 1

Here, we have to find the number of ways in which the 32 students can be seated. We will find the number of ways to seat the 32 students in the two arrangements – first class is seated in row 1, or second class is seated in row 1. Then, we will add the number of ways of the two arrangements to get the total number of ways to seat the 32 students. Since the order of students matter, we will use permutations.

Formula Used: The number of permutations of a set of $n$ objects can be arranged in $r$ places is given by ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$.

Complete step-by-step answer:
We will use two cases to find the number of ways in which the 32 students can be seated.
Let the two classes be class A and class B.
Case 1: The students of class A sit in the first row.
We know that if the students of class A sit in the first row, then the students of class B will have to sit in the second row. Then, the students of class A will sit in the third row and the students of class B will sit in the fourth row.
The arrangement becomes like this: A B A B.
Now, there are 8 chairs in each row and 16 students in each class.
Since the students are distinguishable, we will use permutations instead of combinations.
We know that the number of permutations of a set of $n$ objects can be arranged in $r$ places is given by ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$.
We can observe that out of the 16 students of class A, 8 students need to be seated in row 1.
The number of ways to seat 8 students out of 16 students of class A in row 1 is
${}^{16}{P_8} = \dfrac{{16!}}{{\left( {16 - 8} \right)!}} = \dfrac{{16!}}{{8!}}$
Also, we can observe that out of the 16 students of class B, 8 students need to be seated in row 2.
The number of ways to seat 8 students out of 16 students of class B in row 2 is
${}^{16}{P_8} = \dfrac{{16!}}{{\left( {16 - 8} \right)!}} = \dfrac{{16!}}{{8!}}$
Next, we can observe that out of the remaining 8 students of class A, 8 students need to be seated in row 3.
The number of ways to seat 8 students out of the remaining 8 students of class A in row 3 is
${}^8{P_8} = \dfrac{{8!}}{{\left( {8 - 8} \right)!}} = \dfrac{{8!}}{{0!}} = \dfrac{{8!}}{1} = 8!$
Finally, we can observe that out of the remaining 8 students of class B, 8 students need to be seated in row 4.
The number of ways to seat 8 students out of the remaining 8 students of class B in row 4 is
${}^8{P_8} = \dfrac{{8!}}{{\left( {8 - 8} \right)!}} = \dfrac{{8!}}{{0!}} = \dfrac{{8!}}{1} = 8!$
Now, we will multiply the number of ways to seat the students in the four rows to get the total number of ways to seat the 32 students in the class following the arrangement: A B A B.
Therefore, we get
Number of ways to seat 32 students (A B A B arrangement) $= \dfrac{{16!}}{{8!}} \times \dfrac{{16!}}{{8!}} \times 8!{\rm{ }} \times 8!$
Simplifying the expression, we get
Number of ways to seat 32 students (A B A B arrangement) $= 16!{\rm{ }} \times 16!$
Case 2: The students of class B sit in the first row.
We know that if the students of class B sit in the first row, then the students of class A will have to sit in the second row. Then, the students of class B will sit in the third row and the students of class A will sit in the fourth row.
The arrangement becomes like this: B A B A.
We can observe that out of the 16 students of class B, 8 students need to be seated in row 1.
The number of ways to seat 8 students out of 16 students of class B in row 1 is
${}^{16}{P_8} = \dfrac{{16!}}{{\left( {16 - 8} \right)!}} = \dfrac{{16!}}{{8!}}$
Also, we can observe that out of the 16 students of class A, 8 students need to be seated in row 2.
The number of ways to seat 8 students out of 16 students of class A in row 2 is
${}^{16}{P_8} = \dfrac{{16!}}{{\left( {16 - 8} \right)!}} = \dfrac{{16!}}{{8!}}$
Next, we can observe that out of the remaining 8 students of class B, 8 students need to be seated in row 3.
The number of ways to seat 8 students out of the remaining 8 students of class B in row 3 is
${}^8{P_8} = \dfrac{{8!}}{{\left( {8 - 8} \right)!}} = \dfrac{{8!}}{{0!}} = \dfrac{{8!}}{1} = 8!$
Finally, we can observe that out of the remaining 8 students of class A, 8 students need to be seated in row 4.
The number of ways to seat 8 students out of the remaining 8 students of class A in row 4 is
${}^8{P_8} = \dfrac{{8!}}{{\left( {8 - 8} \right)!}} = \dfrac{{8!}}{{0!}} = \dfrac{{8!}}{1} = 8!$
Now, we will multiply the number of ways to seat the students in the four rows to get the total number of ways to seat the 32 students in the class following the arrangement: B A B A.
Therefore, we get
Number of ways to seat 32 students (B A B A arrangement) $= \dfrac{{16!}}{{8!}} \times \dfrac{{16!}}{{8!}} \times 8!{\rm{ }} \times 8!$
Simplifying the expression, we get
Number of ways to seat 32 students (B A B A arrangement) $= 16!{\rm{ }} \times 16!$
Now, we will add the number of ways to seat the 32 students in the two arrangements to get the total number of ways in which the 32 students can be seated.
Therefore, we get
Number of ways to seat the 32 students $= \left( {16!{\rm{ }} \times {\rm{16!}}} \right) + \left( {16!{\rm{ }} \times {\rm{16!}}} \right) = 2\left( {16!{\rm{ }} \times {\rm{16!}}} \right) = 2{\left( {16!} \right)^2}$
Therefore, we get the total number of ways to seat the 32 students as $2{\left( {16!} \right)^2}$.

Note: It is important for us to note that $0! = 1$. A common mistake we can make is to use $0! = 0$ instead of $0! = 1$as this will give us incorrect answers. Also, we have used permutation to solve the question and not combination because it is asked to find the ways of arrangement where order matters. By the definition, permutation is a way of arranging $r$ objects from $n$ elements where the order or the sequence of arrangement matters. However, combination is a way of selecting $r$ objects from $n$ elements where order doesn’t matter.