The solution to this is to find all possible combinations of male and female to form a team to decorate the party. Of all in attendance, there are 5 males and 3 females. Therefore the team shall consist of;
[tex]\begin{gathered} nC_r=\frac{n!}{r!(n-r)!} \\ \text{Where n is the number of males/females and } \\ r\text{ is the number to be chosen from the total of each category} \\ For\text{ the males;} \\ 5C_1=\frac{5!}{1!(5-1)!} \\ 5C_1=\frac{5\times4\times3\times2\times1}{1\times(4\times3\times2\times1)} \\ 5C_1=\frac{120}{1\times24} \\ 5C_1=5 \\ \text{For the females;} \\ 3C_1=\frac{3!}{1!\times(3-1)!} \\ 3C_1=\frac{3\times2\times1}{1\times(2\times1)} \\ 3C_1=\frac{6}{2} \\ 3C_1=3 \\ \text{Therefore;} \\ 5C_1\times3C_1=5\times3 \\ 5C_1\times3C_1=15 \end{gathered}[/tex]The results shows that they can be selected in 15 different ways
