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Combination

If m and n ar...

Question

If m and n are positive integers greater than or equal to 2, m > n, then (mn)! is divisible by

$(m!)^{n}$

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$(n!)^{m}$

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$(m + n)!$

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$(m - n)!$

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Solution

The correct options are
A
$(m!)^{n}$

B
$(n!)^{m}$

C
$(m + n)!$

D
$(m - n)!$

$\frac{(m n)!}{(m!)^{n}}$ is the number of ways of distributing mn distinct objects in n persons equally. Hence $\frac{(m n)!}{(m!)^{n}}$ is an integer

$\Rightarrow (m!)^{n} | (m n)! .$ Similarly $(n!)^{m} | (m n)!$

Further $m + n < 2 m \leq m n \Rightarrow (m + n)! | (m n)!$ and $m - n < m < m n$

$\Rightarrow (m - n)! | (m n)!$

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