In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

A team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls.

3 boys can be selected from 5 boys in ${ }^{5} \mathrm{C}_{3}$ ways.

3 girls can be selected from 4 girls in ${ }^{4} \mathrm{C}_{3}$ ways.

Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected  $={ }^{5} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{3}=\frac{5 !}{3 ! 2 !} \times \frac{4 !}{3 ! 1 !}$

$=\frac{5 \times 4 \times 3 !}{3 ! \times 2} \times \frac{4 \times 3 !}{3 !}$

$=10 \times 4=40$