Answer:
A)
[tex]\left \{ {{80} \atop {20}} \} * \left \{ {{60} \atop {20}} \} * \left \{ {{40} \atop {20}} \} * \left \{ {{20} \atop {20}} \}[/tex]
B)
[tex]\left \{ {{20+4-1} \atop {4-1}} \} = \left \{ {{23} \atop {3}} \}[/tex]
C)
= [tex]\left \{ {{17+4+1} \atop {4-1}} \} = \left \{ {{20} \atop {3}} \}[/tex]
D)
= [tex]\left \{ {{20} \atop {3}} \} - \left \{ {{17} \atop {3}} \}[/tex]
Step-by-step explanation:
A) How many ways can you put all Jellybeans in a row
Total number of Jellybeans = 80
The first jellybeans = 20 yellow , second is 20 orange jellybeans , third is 20 red jellybeans , fourth is 20 green jellybeans
Therefore the number of ways the Jellybeans can be put in a row is :
[tex]\left \{ {{80} \atop {20}} \} * \left \{ {{60} \atop {20}} \} * \left \{ {{40} \atop {20}} \} * \left \{ {{20} \atop {20}} \}[/tex]
B) How many ways are there to select a handful of 20 jellybeans
lets assume:
yellow jellybeans = a , orange jellybeans = b , red jellybeans = c , green jellybeans = d
a + b + c + d = 20
This is the number Non-negative integer solutions
= [tex]\left \{ {{20+4-1} \atop {4-1}} \} = \left \{ {{23} \atop {3}} \}[/tex]
C) This is also the number of Non-negative integer solutions but in this case the value of C ≥ 3
hence the number of ways to select a handful of 20 jellybeans that contains at least 3 red
= [tex]\left \{ {{17+4+1} \atop {4-1}} \} = \left \{ {{20} \atop {3}} \}[/tex]
D) In this case the value of C ≥ 3 and B ≤ 2
Hence the number of ways to select a handful of 20 jellybeans that contains at least 3 red and at most 2 orange
= [tex]\left \{ {{20} \atop {3}} \} - \left \{ {{17} \atop {3}} \}[/tex]
