Question

# A team consists of $6$ boys and $4$ girls and the other has $5$ boys and $3$ girls. How many single matches can be arranged between the two teams where a boy plays against a boy and girl plays against a girl?

Hint:Take $1$ boy from team $1$and then start counting the no of matches he will play against the boys of team $2$. now take every boy from team $1$ and count the no of matches they will play with the boys of team $2$. Similarly, you can get the total no of matches that girls of team $1$ will play with girls of team $2$.

Let us find the total number of matches that can be arranged between two teams where a boy plays against a boy and girl plays against a girl.
For Boys -
No of boys in team $1$$= 6$
No of boys in team $2$ $= 5$
We have to arrange single matches. This means that every boy from team $1$ will go against every boy of team $2$ .
Hence every boy from team $1$ will play $5$ matches i.e. $1$ match with every boy from team $2$
Since there are $6$ boys in team $1$
Therefore, total no of matches boys will play $6 \times 5 = 30$
For Girls –
No of girls in team $1 = 4$
No of girls in team $2 = 3$
As we did for boys same goes for girls i.e. every girl of team $1$ will compete with every girl of team $2$
Therefore, total no of matches girls will play$= 4 \times 3 = 12$
Total no of matches $=$ matches of boys $+$ matches of girls
$= 30 + 12 = 42$

Note:In this question you just need to take care of the point that every boy of team $1$ will compete with every boy of team $2$ and every girl of team $1$ will compete with every girl of team $2$ . The chances of mistakes anyone could make is while counting the number of matches boys or girls will play . So you need to do that carefully.