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. $({{}^{15}{C}_1)}^2+{{(}^{14}{C}_1)}^2+....+({{}^1{C}_1)}^2$
• B. $15!$
• C. $15{(}^{15}{C}_1{\times }^{15}{C}_1)$
• D. ${}^{15}{C}_1\times 15!$

Answer 1

The correct option is B $15!$

We have to select $15$ teams and every team consists of one man and one woman.

total man $=15$, total woman $=15$

Consider all the $15$ man as group $A$ and all $15$ woman as group $B$,now we have to make $15$ teams taking one member from group $A$ and one member from group $B$.

Now consider one member from group $A$, he has $15$ options to pair up with one member of group $B$,next member from group $A$ has $14$ to pair up with group $B$,third member will have $13$ options and so on.

So the total no. of ways of selecting $15$ teams is
$=15\times 14\times \mathrm{13.....}\times 1=15!$