Question

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

a. How many different ways can four pieces be selected from the 12 colored pieces?
b. How many different ways can two orange pieces be selected from five orange pieces?
c. How many different ways can two brown pieces be selected from seven brown pieces?

Answer 1

a)
`12C4 = (12!)/(4! (12-4)!)= (12!)/(4! 8!)=(12*11*10*9*8!)/(4! 8!)= 495` ways

b)
`5C2 = (5!)/(2! (5-2)!)= (5!)/(2! 3!)=(5*4*3!)/(2! 3!)= 10` ways

c)
`7C2 = (7!)/(2! (7-2)!)= (7!)/(2! 5!)=(7*6*5!)/(2! 5!)= 21` ways

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= (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 = (12!)/(4! (12-4)!)= (12!)/(4! 8!)=(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 = (5!)/(2! (5-2)!)= (5!)/(2! 3!)=(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 = (7!)/(2! (7-2)!)= (7!)/(2! 5!)=(7*6*5!)/(2! 5!)= 21` ways