Answer :
Answer:
There are 91 ways.
Step-by-step explanation:
Selecting five odd OR four even numbers → arrangement is not significant, so we are going to use the Combination method:
[tex]\boxed{_nC_r=\frac{n!}{r!(n-r)!} }[/tex]
where:
- [tex]_nC_r=\text{number of combination}[/tex]
- [tex]n=\text{total number of objects in the set}[/tex]
- [tex]r=\text{number of choosing objects from the set}[/tex]
This question is comprised with 2 events:
- Event A = selecting 5 odd numbers out of 8 odd numbers (1, 3, 5, ..., 15)
- Event B = selecting 4 numbers out of 7 even numbers (2, 4, 6, ..., 14)
(1) Event A:
[tex]\displaystyle _nC_r=\frac{n!}{r!(n-r)!}[/tex]
[tex]\begin{aligned} _8C_5&=\frac{8!}{5!(8-5)!}\\\\&=\frac{8!}{5!3!}\\\\&=\frac{8\times7\times6}{3\times2\times1} \\\\&=56\end{aligned}[/tex]
(2) Event B:
[tex]\displaystyle _nC_r=\frac{n!}{r!(n-r)!}[/tex]
[tex]\begin{aligned} _7C_4&=\frac{7!}{4!(7-4)!}\\\\&=\frac{7!}{4!3!}\\\\&=\frac{7\times6\times5}{3\times2\times1} \\\\&=35\end{aligned}[/tex]
Since both events are mutually exclusive, which means both events do not occur at the same time, then the event A OR event B:
[tex]\boxed{A\ or\ B=A+B}[/tex]
[tex]\begin{aligned}A\ or\ B&=56+35\\&=\bf 91\end{aligned}[/tex]
