Register

I won tickets to a trip for four to the Super Bowl. I have eight friends but I can only bring three of them. In how many ways can I form my Super Bowl party?

I won tickets to a trip for four to the Super Bowl. I have eight friends but I can only bring three of them. In how many ways can I form my Super Bowl party?
151k views
3 votes
I won tickets to a trip for four to the Super Bowl. I have eight friends but I can only bring three of them. In how many ways can I form my Super Bowl party?

User Denise Mauldin
by
8.1k points

2 Answers

4 votes

Answer:

56

Explanation:

User Alexander Smirnov
by
7.7k points
3 votes

8C3=(8!)/(3!5!)=(6\cdot7\cdot8)/(2\cdot3)=56
User Brigette
by
8.7k points
← Prev Question Next Question →

No related questions found

Ask a Question
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories

Other Questions

How do you can you solve this problem 37 + y = 87; y =
What is .725 as a fraction
How do you estimate of 4 5/8 X 1/3
Link Copied!