A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
At a Royal Dynasty Chinese Restaurant, dinner for eight consists of three items from column A, four items from column B, and three items from column C. If columns A, B and C have five, seven, and six items, respectively, how many different dinner combinations are possible?
anonymous
 5 years ago
At a Royal Dynasty Chinese Restaurant, dinner for eight consists of three items from column A, four items from column B, and three items from column C. If columns A, B and C have five, seven, and six items, respectively, how many different dinner combinations are possible?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Separate the question into 3 events. First you choose your items from A, then B, then C. The possibilities in A are 5 choose 3=5!/(2!3!)=10. In B we have 7 choose 4=7!/(3!4!)=35. In C we have 6 choose 3=6!/(3!3!)=20. We multiply the possibilities of each event together to find the total, which is 10*35*20=7000 possibilities.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.