Question

2. Four pieces of candy are drawn at random from a bag containing five orange pieces and seven brown pieces.

Answer 1

Answer:

a) $12C4 = \frac{12!}{4! (12-4)!}= \frac{12!}{4! 8!}=\frac{12*11*10*9*8!}{4! 8!}= 495$ ways

b) $5C2 = \frac{5!}{2! (5-2)!}= \frac{5!}{2! 3!}=\frac{5*4*3!}{2! 3!}= 10$ ways

c) $7C2 = \frac{7!}{2! (7-2)!}= \frac{7!}{2! 5!}=\frac{7*6*5!}{2! 5!}= 21$ 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:

$nCx= \frac{n!}{x! (n-x)!}$

Where $n! = n *(n-1)!$

a. How many different ways can four pieces be selected from the 12 colored pieces?

For this case we have:

$12C4 = \frac{12!}{4! (12-4)!}= \frac{12!}{4! 8!}=\frac{12*11*10*9*8!}{4! 8!}= 495$ ways

b. How many different ways can two orange pieces be selected from five orange pieces?

For this case we have:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5!}{2! 3!}=\frac{5*4*3!}{2! 3!}= 10$ ways

c. How many different ways can two brown pieces be selected from seven brown pieces?

For this case we have:

$7C2 = \frac{7!}{2! (7-2)!}= \frac{7!}{2! 5!}=\frac{7*6*5!}{2! 5!}= 21$ ways