An ice cream parlor has 28 different flavors, 8 different kinds of sauce, and 12 toppings.

A) In how many different ways can a dish of three scoops of ice cream be made where each flavor can be used more than once and the order of the scoops does not matter?
C) How many different kinds of small sundaes are there if a small sundae contains one scoop of ice cream, a sauce, and a topping?
C) How many different kinds of large sundaes are there if a large sundae contains three scoops of ice cream, where each flavor can be used more than once and the order of the scoops does not matter; two kinds of sauce, where each sauce can be used only once and the order of the sauces does not matter; and three toppings, where each topping can be used only once and the order of the toppings does not matter?

(A)Therefore the number of dish of three scoops of ice cream is 4,060.

(B)The different kinds of sundaes are 2668.

(C) Therefore total numbers of different sundaes are 25,009,600.

Step-by-step explanation:

The order of event is counted.

1. Repetition not allowed: P(n,r)  

2. Repetition allowed:

The order of event is not counted.

1. Repetition not allowed:

2. Repetition allowed:

(A)

The order of scope does not matter and repetition allowed. So we use the second formula of combination.

The scoops are then 3 section from 28 different flavors

n=28

r=3

Therefore the number of dish of three scoops of ice cream is 4,060.

(B)

Given a small sundae contains one scoop of ice cream, a sauce and a topping.

The different kinds of sundaes are C(28,1)×C(8,1)×C(12,1)

                                                           =28×8×12

                                                           =2688

(C)

The order of the scoop does not matter and repetition is allowed.

So, we use the definition of combination of repetition.

n=28

r=3

The order of the sauce does not matter and repetition is not allowed.

So, we use the definition of combination of repetition is not allowed.

n=8

r=2

The order of the topping does not matter and repetition is not allowed.

So, we use the definition of combination of repetition is not allowed.

n=12

r=3

Therefore total number of different sundaes are = (4060×28×220)

                                                                                    =25,009,600

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer:   (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

only 10/60 isn't equivalent

hope this : )