Question

The number of ways of selecting $15$ teams from $15$ men and $15$ women such that each team consists of a man and a woman, is

• A. $14!$
• B. $15!$
• C. $(15!{)}^2$
• D. $1960$

Answer 1

The correct option is B $15!$

Teams of two have to be made, each consisting of $1$ man and $1$ woman.
Let us fix the man in each team.
A woman can be chosen for the first team in $15$ ways.
A woman can be chosen for the second team in $14$ ways.
$⋮$
A woman can be chosen for the ${15}^{th}$ team in $1$ way.
Hence, required number of ways $=15\times 14\times \cdots \times 1=15!$