Solve the following problems completely. Write the answer before the number.
6. Rylan's basketball team has 11 players. How many ways can his
coach choose five starting players?
7. Ashlyn needs to bring a dessert to her dinner party. If the bakery has
eight pies to choose from, how many ways can Ashlyn choose three?
8. On a circle there are 11 points selected. How many triangles with
edges in these points exist?
9. How many combinations can the seven colors of the rainbow be
arranged into groups of two colors each?
10. How many groups can be made from the word "house" if each group consists of 4 alphabets?
Answers & Comments
Answer:
Rylan's basketball team has 11 players. How many ways can his coach choose five starting players?
C(n,r) = (n!)/(r!(n-r)!)
C(n,r) = (n!)/(r!(n-r)!)C(11,5) = (11!)/(5!*(11-5)!)
C(n,r) = (n!)/(r!(n-r)!)C(11,5) = (11!)/(5!*(11-5)!)C(11,5) = (11!)/(5!*6!)
C(n,r) = (n!)/(r!(n-r)!)C(11,5) = (11!)/(5!*(11-5)!)C(11,5) = (11!)/(5!*6!)C(11,5) = (11*10*9*8*7*6!)/(5!*6!)
C(n,r) = (n!)/(r!(n-r)!)C(11,5) = (11!)/(5!*(11-5)!)C(11,5) = (11!)/(5!*6!)C(11,5) = (11*10*9*8*7*6!)/(5!*6!)C(11,5) = (11*10*9*8*7)/(5!)
C(n,r) = (n!)/(r!(n-r)!)C(11,5) = (11!)/(5!*(11-5)!)C(11,5) = (11!)/(5!*6!)C(11,5) = (11*10*9*8*7*6!)/(5!*6!)C(11,5) = (11*10*9*8*7)/(5!)C(11,5) = (11*10*9*8*7)/(5*4*3*2*1)
C(n,r) = (n!)/(r!(n-r)!)C(11,5) = (11!)/(5!*(11-5)!)C(11,5) = (11!)/(5!*6!)C(11,5) = (11*10*9*8*7*6!)/(5!*6!)C(11,5) = (11*10*9*8*7)/(5!)C(11,5) = (11*10*9*8*7)/(5*4*3*2*1)C(11,5) = (55440)/(120)
C(n,r) = (n!)/(r!(n-r)!)C(11,5) = (11!)/(5!*(11-5)!)C(11,5) = (11!)/(5!*6!)C(11,5) = (11*10*9*8*7*6!)/(5!*6!)C(11,5) = (11*10*9*8*7)/(5!)C(11,5) = (11*10*9*8*7)/(5*4*3*2*1)C(11,5) = (55440)/(120)C(11,5) = 462
C(n,r) = (n!)/(r!(n-r)!)C(11,5) = (11!)/(5!*(11-5)!)C(11,5) = (11!)/(5!*6!)C(11,5) = (11*10*9*8*7*6!)/(5!*6!)C(11,5) = (11*10*9*8*7)/(5!)C(11,5) = (11*10*9*8*7)/(5*4*3*2*1)C(11,5) = (55440)/(120)C(11,5) = 462So there are 462 ways.