Respuesta :
Answer:
The number of ways those textbooks could be distributed among the students is expressed by a combinations of 8(books) to 5(students), which is [tex]C_{8,5} = \frac{8!}{5!(8-5)!} = 56[/tex]
Step-by-step explanation:
The order in which the students get the textbooks is not important. For example, A,B,C,D and E getting the textbooks is the same outcome as B,A,C,D and E. So we use the combinations formula to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem
5 textbooks to 8 students. So
[tex]C_{8,5} = \frac{8!}{5!(8-5)!} = 56[/tex]
The number of ways those textbooks could be distributed among the students is expressed by a combinations of 8(books) to 5(students), which is [tex]C_{8,5} = \frac{8!}{5!(8-5)!} = 56[/tex]
