Q1=The number of ways in which 2m distinct things can be divided into 2 groups so that they contain m things each is?
Q2=The number of ways in which 2m distinct things can be divided into 2 persons so that they contain m things each is?

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Solution

1) The number of ways in which 2m distinct things can be divided into 2 groups so that they contain m things each is
$\frac{(2 m)!}{m! m! 2!}$ , here 2! is in the denominator because m things are interchanged into both groups

2)The number of ways in which 2m distinct things can be divided into 2 persons so that they contain m things each is
$\frac{(2 m)!}{m! m!}$

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Q1=The number of ways in which 2m distinct things can be divided into 2 groups so that they contain m things each is?Q2=The number of ways in which 2m distinct things can be divided into 2 persons so that they contain m things each is?

Q1=The number of ways in which 2m distinct things can be divided into 2 groups so that they contain m things each is?
Q2=The number of ways in which 2m distinct things can be divided into 2 persons so that they contain m things each is?