Question

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?

Answer 1

A) There are 3276 different ways to make a dish of three scoops of ice cream; B) There are 2688 different kinds of small sundaes; C) There are 20,154,240 different kinds of large sundaes.

Step-by-step explanation:

Let's solve each part of the question one by one:

A) In this case, we need to find the number of ways we can select 3 scoops of ice cream from 28 flavors where each flavor can be used more than once and the order of the scoops does not matter. This is a combination problem because the order does not matter. We can solve it using the formula for combinations:

nCr = n! / (r! * (n-r)!)

Plugging in the values, we get:

28C3 = 28! / (3! * (28-3)!) = 3276

So, there are 3276 different ways to make a dish of three scoops of ice cream.

B) In this case, a small sundae contains one scoop of ice cream, a sauce, and a topping. We need to find the number of combinations. There are 28 flavors of ice cream, 8 kinds of sauce, and 12 toppings. Using the formula for combinations, we get:

28C1 * 8C1 * 12C1 = 28 * 8 * 12 = 2688

So, there are 2688 different kinds of small sundaes.

C) In this case, a large sundae contains three scoops of ice cream, two kinds of sauce, and three toppings. Again, using the formula for combinations, we get:

28C3 * 8C2 * 12C3 = 3276 * 28 * 220 = 20154240

So, there are 20,154,240 different kinds of large sundaes.

Answer 2

(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:

• Definition of permutation:

The order of event is counted.

1. Repetition not allowed: P(n,r)
`=(n!)/((n-r)!)`

2. Repetition allowed:
`n^r`

• Definition of combination

The order of event is not counted.

1. Repetition not allowed:

`C(n,r) =\left(\begin{array}{ccc}n\\r\end{array}\right)=(n!)/(r!(n-r)!)`

2. Repetition allowed:

`C(n+r-1,r) =\left(\begin{array}{ccc}n+r-1\\r\end{array}\right)=((n+r-1)!)/(r!(n-1)!)`

(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 C(28+3-1,3) =\left(\begin{array}{c}30\\3\end{array}\right)=((30)!)/(3!27!)=4,060`

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)

Scoop:

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

`C(28+3-1,3)=C(30,3)=(30!)/(3!27!)=4060`

Sauce:

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

`C(8,2)=(8!)/(2!(8-2)!)=(8!)/(2!6!)=28`

Topping:

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

`C(12,3)=(12!)/(3!(12-3)!)=(12!)/(3!9!)=220`

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

=25,009,600