The number of ways in which 2n different things can be divided equally into two distinct groups is $\frac{2 n!}{2 (n!)^{2}}$ (If order of the group is NOT important)

True

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False

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Solution

The correct option is A
True

If order of the group is NOT important
The number of ways in which 2n different things can be divided equally into two distinct groups is = $\frac{2 n!}{2! \times n! \times n!} = \frac{2 n!}{2! \times (n!)^{2}}$

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The number of ways in which 2n different things can be divided equally into two distinct groups is 2n!2(n!)2 (If order of the group is NOT important)

The correct option is A True If order of the group is NOT important The number of ways in which 2n different things can be divided equally into two distinct groups is =2n!2!×n!×n!=2n!2!×(n!)2

The number of ways in which 2n different things can be divided equally into two distinct groups is $\frac{2 n!}{2 (n!)^{2}}$ (If order of the group is NOT important)

The correct option is A
True

If order of the group is NOT important
The number of ways in which 2n different things can be divided equally into two distinct groups is = $\frac{2 n!}{2! \times n! \times n!} = \frac{2 n!}{2! \times (n!)^{2}}$