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Found in: Page 440

### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

# An ice cream parlour has $${\rm{28}}$$ different flavours, $${\rm{8}}$$ different kinds of sauce, and $${\rm{12}}$$ toppings. a) In how many different ways can a dish of three scoops of ice cream be made where each flavour can be used more than once and the order of the scoops does not matter? b) 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 flavour 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) The number of different ways a dish of three scoops of ice cream can be made where each flavour can be used more than once and the order of the scoops does not matter is $${\rm{4060}}$$.

(b) The number of different kinds of small sundaes there are if a small sundae contains one scoop of ice cream, a sauce, and a topping is $${\rm{2688}}$$.

(c) The number of different kinds of large sundaes there are if a large sundae contains three scoops of ice cream is $${\rm{25,009,600}}$$.

See the step by step solution

## Step 1: Concept Introduction

Product rule: If one event can occur in $${\rm{m}}$$ ways and a second event can occur in $${\rm{n}}$$ ways, then the number of ways that the two events can occur in sequence is then $${\rm{m}} \cdot {\rm{n}}$$.

Definition of permutation (order is important) is –

No repetition allowed: $${\rm{P(n,r) = }}\frac{{{\rm{n!}}}}{{{\rm{(n - r)!}}}}$$

Repetition allowed: $${{\rm{n}}^{\rm{r}}}$$

Definition of combination (order is important) is –

No repetition allowed: $${\rm{C(n,r) = }}\frac{{{\rm{n!}}}}{{{\rm{r!(n - r)!}}}}$$

Repetition allowed: $${\rm{C(n + r - 1,r) = }}\frac{{{\rm{(n + r - 1)!}}}}{{{\rm{r!(n - 1)!}}}}$$

With $${\rm{n! = n}} \cdot {\rm{(n - 1)}} \cdot ... \cdot {\rm{2}} \cdot {\rm{1}}$$.

## Step 2: Finding the number of different ways

(a)

The order of the scoops does not matter; thus, it is needed to use the definition of combination.

The scoops are then $${\rm{3}}$$ selections from the $${\rm{28}}$$ different flavours.

Here, it can be seen $${\rm{n = 28, r = 3}}$$.

Repetition is allowed (since each flavour can be used more than once), so substitute the value and calculate –

$$\begin{array}{c}{\rm{C(n + r - 1,r) = C(28 + 3 - 1,103)}}\\{\rm{ = C(30,3)}}\\{\rm{ = }}\frac{{{\rm{30!}}}}{{{\rm{3!(30 - 3)!}}}}\\{\rm{ = }}\frac{{{\rm{30!}}}}{{{\rm{3!27!}}}}\\{\rm{ = 4060}}\end{array}$$

Therefore, the result is obtained as $${\rm{4060}}$$.

## Step 3: Finding the different kinds of small sundaes

(b)

It is needed to choose one scoop, one sauce and one topping –

There are –

Scoops: $${\rm{28}}$$ ways (flavours)

Sauce: $${\rm{8}}$$ ways

Toppings: $${\rm{12}}$$ ways

Using the product rule –

$${\rm{28}} \cdot 8 \cdot {\rm{12 = 2688}}$$

Therefore, the result is obtained as $${\rm{2688}}$$.

## Step 4: Finding the different kinds of large sundaes

(c)

First consider the calculation for scoops.

The order of the scoops does not matter; thus, it is needed to use the definition of combination.

The scoops are then $${\rm{3}}$$ selections from the $${\rm{28}}$$ different flavours.

Here, it can be seen $${\rm{n = 28, r = 3}}$$.

Repetition is allowed (since each flavour can be used more than once), so substitute the value and calculate –

$$\begin{array}{c}{\rm{C(n + r - 1,r) = C(28 + 3 - 1,103)}}\\{\rm{ = C(30,3)}}\\{\rm{ = }}\frac{{{\rm{30!}}}}{{{\rm{3!(30 - 3)!}}}}\\{\rm{ = }}\frac{{{\rm{30!}}}}{{{\rm{3!27!}}}}\\{\rm{ = 4060}}\end{array}$$

Now consider the calculation for sauces.

The order of the sauces does not matter; thus, it is needed to use the definition of combination.

The scoops are then $${\rm{2}}$$ selections from the $${\rm{8}}$$ different flavours.

Here, it can be seen $${\rm{n = 8, r = 2}}$$.

Repetition is not allowed (since each sauce can be used only once), so substitute the value and calculate –

$$\begin{array}{c}{\rm{C(8,2) = }}\frac{{{\rm{8!}}}}{{{\rm{2!(8 - 2)!}}}}\\{\rm{ = }}\frac{{{\rm{8!}}}}{{{\rm{2!6!}}}}\\{\rm{ = 28}}\end{array}$$

Now consider the calculation for toppings.

The order of the toppings does not matter; thus, it is needed to use the definition of combination.

The scoops are then $${\rm{3}}$$ selections from the $${\rm{12}}$$ different flavours.

Here, it can be seen $${\rm{n = 12, r = 3}}$$.

Repetition is not allowed (since each topping can be used only once), so substitute the value and calculate –

$$\begin{array}{c}{\rm{C(12,3) = }}\frac{{{\rm{12!}}}}{{{\rm{3!(12 - 3)!}}}}\\{\rm{ = }}\frac{{{\rm{12!}}}}{{{\rm{3!9!}}}}\\{\rm{ = 220}}\end{array}$$

Using the product rule –

$${\rm{4060}} \cdot 28 \cdot {\rm{220 = 25,009,600}}$$

Therefore, the result is obtained as $${\rm{25,009,600}}$$.