The number of ways in which 2n different things can be divided equally into two distinct groups is $\frac{2 n!}{(n!)^{2}}$ (If order of the group is important)

True

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False

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Solution

The correct option is A
True

If order of the group is important
The number of ways in which 2n different things can be divided equally into two distinct groups is
$\frac{2 n!}{2! \times n! \times n!} \times 2! = \frac{2 n!}{(n!)^{2}}$

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The number of ways in which 2n different things can be divided equally into two distinct groups is 2n!(n!)2 (If order of the group is important)

The correct option is A True If order of the group is important The number of ways in which 2n different things can be divided equally into two distinct groups is 2n!2!×n!×n!×2!=2n!(n!)2

The number of ways in which 2n different things can be divided equally into two distinct groups is $\frac{2 n!}{(n!)^{2}}$ (If order of the group is important)

The correct option is A
True

If order of the group is important
The number of ways in which 2n different things can be divided equally into two distinct groups is
$\frac{2 n!}{2! \times n! \times n!} \times 2! = \frac{2 n!}{(n!)^{2}}$