Question

The number of ways in which we can distribute mn students equally among m sections is given by:
(a) \[\dfrac{\left( mn \right)!}{n!}\]
(b) \[\dfrac{\left( mn \right)!}{{{\left( n! \right)}^{m}}}\]
(c) \[\dfrac{\left( mn \right)!}{m!n!}\]
(d) \[{{\left( mn \right)}^{m}}\]

Answer 1

Hint: In order to solve this question, we should have some knowledge regarding the concept of combination, that is for choosing r out of n items irrespective of their order we apply the formula, \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]. And therefore, we will choose n number of students for each section turn by turn.

Complete step-by-step answer:
In this question, we have been asked to find the number of ways in which we can distribute mn students equally among m section. Now, we have been given that there are mn students in total and they have to be distributed equally among m sections. So, we can say in each section, there will be \[\dfrac{mn}{m}=n\] students.
Now, we take n students from mn students for each section by using the formula of combination, that is,
\[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
For the first section, we can choose n students out of mn. So, we get the number of ways of choosing n students for the first section as \[^{mn}{{C}_{n}}\].
For the second section, we will again choose n students but out of (mn – n) because n students have already been chosen for the first section. So, we get the number of ways of choosing n students for the second section as \[^{mn-n}{{C}_{n}}\].
Similarly, for the third section, we have to choose n students out of (mn – 2n). So, we get the number of ways of choosing n students for the third section as \[^{mn-2n}{{C}_{n}}\].
And we will continue it in the same manner up to all mn students will not be divided into m section.
So, for (m – 1)th section, we will choose n students from (mn – ( m – 2)n) student. So, we get the number of ways of choosing n students for (n – 1)th section, we get, \[^{mn-\left( m-2 \right)n}{{C}_{n}}\].
And for the mth section, we get the number of ways for choosing students as, \[^{mn-\left( m-1 \right)n}{{C}_{n}}\].
Hence, we can write the total number of ways of distributing mn students in m section as
\[^{mn}{{C}_{n}}{{\times }^{mn-n}}{{C}_{n}}{{\times }^{mn-2n}}{{C}_{n}}\times ....{{\times }^{mn-\left( m-2 \right)n}}{{C}_{n}}{{\times }^{mn-\left( m-1 \right)n}}{{C}_{n}}\]
Now, we will use the formula of \[^{n}{{C}_{r}}\] to expand it. So, we get,
\[\dfrac{\left( mn \right)!}{n!\left( mn-n \right)!}\times \dfrac{\left( mn-n \right)!}{n!\left( mn-2n \right)!}\times \dfrac{\left( mn-2n \right)!}{n!\left( mn-3m \right)!}\times .....\times \dfrac{\left( mn-\left( m-2 \right)n \right)!}{n!\left( mn-\left( m-1 \right)n \right)!}\times \dfrac{\left( mn-\left( m-1 \right)n \right)!}{n!\left( mn-mn \right)!}\]
And we can further write it as,
\[\dfrac{\left( mn \right)!}{n!}\times \dfrac{1}{n!}\times \dfrac{1}{n!}\times .....\times 1\]
\[\dfrac{\left( mn \right)!}{{{\left( n! \right)}^{m}}}\]
Hence, we can say that the total number of ways of distributing mn students in m section are \[\dfrac{\left( mn \right)!}{{{\left( n! \right)}^{m}}}\].
Therefore, option (a) is the right answer.

Note: While solving this question, the possible mistake one can make is by always choosing n students for all sections from mn students which is totally wrong because at a time one student can only be in 1 section. So, if n students are selected for 1 section then in the second section, we will choose from (mn – n).