Respuesta :
Answer:
a) [tex] 12C4 = \frac{12!}{4! (12-4)!}= \frac{12!}{4! 8!}=\frac{12*11*10*9*8!}{4! 8!}= 495[/tex] ways
b) [tex] 5C2 = \frac{5!}{2! (5-2)!}= \frac{5!}{2! 3!}=\frac{5*4*3!}{2! 3!}= 10[/tex] ways
c) [tex] 7C2 = \frac{7!}{2! (7-2)!}= \frac{7!}{2! 5!}=\frac{7*6*5!}{2! 5!}= 21[/tex] ways
Step-by-step explanation:
Combinatory means combination or arrangement of different elements.
If we have n total elements and we want to find in how many ways we can select x we can use this general formula:
[tex] nCx= \frac{n!}{x! (n-x)!}[/tex]
Where [tex] n! = n *(n-1)![/tex]
a. How many different ways can four pieces be selected from the 12 colored pieces?
For this case we have:
[tex] 12C4 = \frac{12!}{4! (12-4)!}= \frac{12!}{4! 8!}=\frac{12*11*10*9*8!}{4! 8!}= 495[/tex] ways
b. How many different ways can two orange pieces be selected from five orange pieces?
For this case we have:
[tex] 5C2 = \frac{5!}{2! (5-2)!}= \frac{5!}{2! 3!}=\frac{5*4*3!}{2! 3!}= 10[/tex] ways
c. How many different ways can two brown pieces be selected from seven brown pieces?
For this case we have:
[tex] 7C2 = \frac{7!}{2! (7-2)!}= \frac{7!}{2! 5!}=\frac{7*6*5!}{2! 5!}= 21[/tex] ways
