The number of ways in which mn students can be distributed equally among n sections is-

(m n)^{n}

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\frac{(m n)!}{(m!)^{n}}

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\frac{(m n)!}{m!}

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\frac{(m n)!}{(m!)^{n} n!}

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Solution

The correct option is C $\frac{(m n)!}{(m!)^{n}}$
The no. of ways in which $m n$ students can be distributed equally among $n$ section is,

$No. of ways in which m \times n different objects can be distributed.$

$= (No. of ways of dividing) \times (no. of groups)!$

$= (m n! \times \frac{n!}{{m!}^{n}}) = \frac{(m n)!}{{m!}^{n}}$

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The number of ways in which mn students can be distributed equally among n sections is-

The correct option is C (mn)!(m!)nThe no. of ways in which mn students can be distributed equally among n section is,No. of ways in which m×n different objects can be distributed.=(No. of ways of dividing)×(no. of groups)!=(mn!×n!m!n)=(mn)!m!n