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Question: The number of ways in which we can distribute mn students equally among m sections is given by: (a...

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

Explanation

Solution

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, nCr=n!r!(n−r)!^{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 mnm=n\dfrac{mn}{m}=n students.
Now, we take n students from mn students for each section by using the formula of combination, that is,
nCr=n!r!(n−r)!^{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 mnCn^{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−nCn^{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−2nCn^{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−(m−2)nCn^{mn-\left( m-2 \right)n}{{C}_{n}}.
And for the mth section, we get the number of ways for choosing students as, mn−(m−1)nCn^{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
mnCn×mn−nCn×mn−2nCn×....×mn−(m−2)nCn×mn−(m−1)nCn^{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 nCr^{n}{{C}_{r}} to expand it. So, we get,
(mn)!n!(mn−n)!×(mn−n)!n!(mn−2n)!×(mn−2n)!n!(mn−3m)!×.....×(mn−(m−2)n)!n!(mn−(m−1)n)!×(mn−(m−1)n)!n!(mn−mn)!\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,
(mn)!n!×1n!×1n!×.....×1\dfrac{\left( mn \right)!}{n!}\times \dfrac{1}{n!}\times \dfrac{1}{n!}\times .....\times 1
(mn)!(n!)m\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 (mn)!(n!)m\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).