Register
A group of 8 friends (5 girls and 3 boys) plans to watch a movie, but they have only 5 tickets. How many different combinations of 5 friends could possibly receive the tickets? A. 13 B. 40 C. 56 b. 64​
129k views
4 votes
A group of 8 friends (5 girls and 3 boys) plans to watch a movie, but they have only 5 tickets. How many different combinations of 5 friends could

possibly receive the tickets?
A. 13
B. 40
C. 56
b. 64​

User Xtian Macedo
by
4.6k points

1 Answer

4 votes


_8C_5=(8!)/(5!3!)=(6\cdot7\cdot8)/(2\cdot 3)=56

User HV Sharma
by
4.5k points
Ask a Question
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

5.1m questions

6.7m answers

Other Questions

Pizza planet is running a special 3 pizzas for 16.50. What is the unit rate for one pizza
What number is halfway between 0 and 18
If the 9-inch wheel of cheese costs $18.60, what is the cost per square inch? If the cost is less than a dollar, put a zero to the left of the decimal point?
I need to simplify this expression.
Need answer to math problem!!!​
Link Copied!