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# 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?A. $2 \times 16! \times 16!$B. $4 \times 16! \times 16!$C. $2 \times 8! \times 16!$D. None of these

Last updated date: 02nd Aug 2024
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Hint: We solve this question by proper visualization and arrangements of provided data and use the formula of permutation to find the number of ways to seat 32 students.
* Permutation is an arrangement of all or a part of a set of objects keeping in mind the order of the selection. Number of ways n people can be seated in r seats is $^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$

Two classes having 32 people in total can be arranged in 4 rows having 8 chairs each, and will be in the following 2 ways in Row1 Row2 Row3 Row4 respectively.
${1^{st}}$way: Class1 Class2 Class1 Class2 or
${2^{nd}}$way: Class2 Class1 Class2 Class1
Total arrangements = Total arrangements in ${1^{st}}$ way+ Total arrangements in the${2^{nd}}$ way
We know from the formula $^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$, so if n=r then $^n{P_n} = \dfrac{{n!}}{{(n - n)!}} = \dfrac{{n!}}{{0!}} = n!$
Since we know each class has total number of students as 16, so
Number of arrangements made to seat 16 people of class1 in 16 chairs ${ = ^{16}}{P_{16}} = 16!$
Number of arrangements made to seat 16 people of class2 in 16 chairs ${ = ^{16}}{P_{16}} = 16!$
Arrangements in ${1^{st}}$ way =$16! \times 16!$
Arrangements in${2^{nd}}$ way =$16! \times 16!$
Hence total arrangement = $16! \times 16! + 16! \times 16!$ $= 2 \times 16! \times 16!$

So, the correct answer is “Option A”.

Note: Students might make mistakes in writing $0! = 0$ which is wrong because it has the value 1. Students might get confused whether to apply combination formula or permutation formula but keep in mind that permutation is used when the order of selection matters and combination is used when order of selection does not matter. Permutation and Combination is one tricky topic where application of it makes confusions while solving.