Statement-1: If n is a natural number thenn2!(n!)n+1 is a natural number.Statement-2: The number of ways of dividing mn studentsinto m groups each containing n students is (mn)!m!(n!)m

A
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for
STATEMENT-1
C
STATEMENT-1 is True, STATEMENT-2 is False
D
STATEMENT-1 is False, STATEMENT-2 is True

Solution:

The number of ways of selecting students for
the first group is $^{m n} C_{n}$ ; for the second group is $^{m n - n} C_{n}$ and

so on.

$∴$ the number of ways of dividing $(m n)$ students into m
numbered groups is

$\begin{array}{l} (^{m n} C_{n}) (^{m n - n} C_{n}) \dots (^{n} C_{n}) \\ = \frac{(m n)!}{n! (m n - n)!} \frac{(m n - n)!}{n! (m n - 2 n)!} \dots \frac{n!}{n! 0!} \\ = \frac{(m n)!}{(n!)^{m}} \end{array}$

Statement-1: If n is a natural number then
$\frac{(n^{2})!}{(n!)^{n + 1}}$ is a natural number.
Statement-2: The number of ways of dividing mn students
into m groups each containing n students is $\frac{(m n)!}{m! (n!)^{m}}$

Solution:

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