Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
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]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.