Question

A croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccoli croissants. How many ways are there to choose

a) a dozen croissants?

b) three dozen croissants?

c) two dozen croissants with at least two of each kind?

d) two dozen croissants with no more than two broccoli croissants?

e) two dozen croissants with at least five chocolate croissants and at least three almond croissants?

f ) two dozen croissants with at least one plain croissant, at least two cherry croissants, at least three chocolate croissants, at least one almond croissant, at least two apple croissants, and no more than three broccoli croissants?

Short answer

There are \(6188\) different ways to choose a dozen croissants.

b) There are \(749,398\) different ways to choose three dozen croissants.

c) There are \(6188\) different ways to choose two dozen croissants with at least two of each kind.

d) There are \(52,975\) different ways to choose two dozen croissants with no more than two broccoli croissants

e) There are \(20,349\) different ways to choose two dozen croissants with at least five chocolate croissants and at least three almond croissants

f) There are \(11136\) different ways of choosing two dozen croissants with at least one plain croissant, at least two cherry croissants, at least three chocolate croissants, at least one almond croissant, at least two apple croissants, and no more than three broccoli croissants.

Step-by-step solution

Step 1(a): Definitions

Definition of Permutation (Order is important)

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

Repetition allowed:\({n^r}\)

Definition of combination (order is important)

No repetition allowed:\(C(n,r) = \left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right) = \frac{{n!}}{{r!(n - r)!}}\)

Repetition allowed:\(C(n + r - 1,r) = \left( {\begin{array}{*{20}{c}}{n + r - 1}\\r\end{array}} \right) = \frac{{(n + r - 1)!}}{{r!(n - 1)!}}\)

with\(n! = n(n - 1) \cdot ..... \cdot 2 \cdot 1\)

Step 2(a): Solution

The order of the elements does not matters (given), thus we need to use the definition of combination.

The shop has six types of croissants (plain, cherry, chocolate, almond, apple, broccoli) of which twelve (a dozen) have to be selected.

We are interested in selecting \(r = 12\)elements from a set with \(n = 6\)elements.

Repetition of elements is allowed

\(\begin{array}{l}C(n + r - 1,r) = C(6 + 12 - 1,12)\\C(17,12) = \frac{\begin{array}{l}\\17!\end{array}}{{12!(17 - 12)!}} = \frac{{17!}}{{12!5!}} = 6188\end{array}\)

Step 1(b): Definitions

Definition of Permutation (Order is important)

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

Repetition allowed:\({n^r}\)

Definition of combination (order is important)

No repetition allowed:\(C(n,r) = \left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right) = \frac{{n!}}{{r!(n - r)!}}\)

Repetition allowed:\(C(n + r - 1,r) = \left( {\begin{array}{*{20}{c}}{n + r - 1}\\r\end{array}} \right) = \frac{{(n + r - 1)!}}{{r!(n - 1)!}}\)

with\(n! = n(n - 1) \cdot ..... \cdot 2 \cdot 1\)

Step 2(b): Solution

The order of the elements does not matters (given), thus we need to use the definition of combination.

The shop has six types of croissants (plain, cherry, chocolate, almond, apple, broccoli) of which thirty-six (three dozen) have to be selected.

We are interested in selecting \(r = 36\)elements from a set with \(n = 6\)elements.

Repetition of elements is allowed

\(\begin{array}{l}C(n + r - 1,r) = C(6 + 36 - 1,36)\\C(41,36) = \frac{\begin{array}{l}\\41!\end{array}}{{36!(41 - 36)!}} = \frac{{41!}}{{36!5!}} = 749,398\end{array}\)

Step 1(c): Definitions

Definition of Permutation (Order is important)

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

Repetition allowed:\({n^r}\)

Definition of combination (order is important)

No repetition allowed:\(C(n,r) = \left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right) = \frac{{n!}}{{r!(n - r)!}}\)

Repetition allowed:\(C(n + r - 1,r) = \left( {\begin{array}{*{20}{c}}{n + r - 1}\\r\end{array}} \right) = \frac{{(n + r - 1)!}}{{r!(n - 1)!}}\)

with\(n! = n(n - 1) \cdot ..... \cdot 2 \cdot 1\)

Step 2(c): Solution

The order of the elements does not matters (given), thus we need to use the definition of combination.

The shop has six types of croissants (plain, cherry, chocolate, almond, apple, broccoli) of which twenty-four (two dozen) have to be selected.

Let us first select two of each kind, which are twelve croissant in total. Then we still need to select the remaining twelve croissants.

We are interested in selecting \(r = 12\)elements from a set with \(n = 6\)elements.

Repetition of elements is allowed

\(\begin{array}{l}C(n + r - 1,r) = C(6 + 12 - 1,12)\\C(17,12) = \frac{\begin{array}{l}\\17!\end{array}}{{12!(17 - 12)!}} = \frac{{17!}}{{12!5!}} = 6188\end{array}\)

Step 1(d): Definitions

Definition of Permutation (Order is important)

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

Repetition allowed:\({n^r}\)

Definition of combination (order is important)

No repetition allowed:\(C(n,r) = \left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right) = \frac{{n!}}{{r!(n - r)!}}\)

Repetition allowed:\(C(n + r - 1,r) = \left( {\begin{array}{*{20}{c}}{n + r - 1}\\r\end{array}} \right) = \frac{{(n + r - 1)!}}{{r!(n - 1)!}}\)

with\(n! = n(n - 1) \cdot ..... \cdot 2 \cdot 1\)

Step 2(d): No broccoli

The order of the elements does not matters (given), thus we need to use the definition of combination.

The shop has six types of croissants (plain, cherry, chocolate, almond, apple, broccoli) of which twenty-four (two dozen) have to be selected.

But the condition is no broccoli so now we assume that the shop has only five croissants.

We are interested in selecting \(r = 24\) (two dozen) elements from a set with \(n = 5\)elements.

Repetition of elements is allowed

\(\begin{array}{l}C(n + r - 1,r) = C(5 + 24 - 1,24)\\C(28,24) = \frac{\begin{array}{l}\\28!\end{array}}{{24!(28 - 24)!}} = \frac{{28!}}{{24!4!}} = 20,475\end{array}\)

Step 3(d): One broccoli

Now for one broccoli croissant

The shop has six types of which twenty-four have to be selected. Let us first select one broccoli croissant. Then we still need to select remaining 23 croissants from the five types of croissants remaining.

We are interested in selecting \(r = 23\)elements from a set with \(n = 5\)elements.

\(\begin{array}{l}C(n + r - 1,r) = C(5 + 23 - 1,23)\\C(27,23) = \frac{\begin{array}{l}\\27!\end{array}}{{23!(27 - 23)!}} = \frac{{27!}}{{23!4!}} = 17,550\end{array}\)

Step 4(d): Two broccoli

Now for two broccoli croissant

The shop has six types of which twenty-four have to be selected. Let us first select two broccoli croissant. Then we still need to select remaining 22 croissants from the five types of croissants remaining.

We are interested in selecting \(r = 22\)elements from a set with \(n = 5\)elements.

\(\begin{array}{l}C(n + r - 1,r) = C(5 + 22 - 1,22)\\C(26,22) = \frac{\begin{array}{l}\\26!\end{array}}{{22!(26 - 22)!}} = \frac{{26!}}{{22!4!}} = 14,950\end{array}\)

Step 5(d): The sum rule

Now as we have all the conditions we need to add them all to get the final result.

So from step two, three and four we get:

\(20475 + 17550 + 14950 = 52,975\)

Step 1(e): Definitions

Definition of Permutation (Order is important)

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

Repetition allowed:\({n^r}\)

Definition of combination (order is important)

No repetition allowed:\(C(n,r) = \left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right) = \frac{{n!}}{{r!(n - r)!}}\)

Repetition allowed:\(C(n + r - 1,r) = \left( {\begin{array}{*{20}{c}}{n + r - 1}\\r\end{array}} \right) = \frac{{(n + r - 1)!}}{{r!(n - 1)!}}\)

with\(n! = n(n - 1) \cdot ..... \cdot 2 \cdot 1\)

Step 2(e): Solution

The order of the elements does not matters (given), thus we need to use the definition of combination.

The shop has six types of croissants (plain, cherry, chocolate, almond, apple, broccoli) of which twenty-four (two dozen) have to be selected.

Let us first select five chocolate croissants and three almond croissant. Then we still need to select the remaining \(24 - 5 - 3 = 16\;croissants\)

We are interested in selecting \(r = 16\) elements from a set with \(n = 6\)elements.

Repetition of elements is allowed

\(\begin{array}{l}C(n + r - 1,r) = C(6 + 16 - 1,16)\\C(21,16) = \frac{\begin{array}{l}\\21!\end{array}}{{16!(21 - 16)!}} = \frac{{21!}}{{16!5!}} = 20,349\end{array}\)

Step 1(f): Definitions

Definition of Permutation (Order is important)

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

Repetition allowed:\({n^r}\)

Definition of combination (order is important)

No repetition allowed:\(C(n,r) = \left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right) = \frac{{n!}}{{r!(n - r)!}}\)

Repetition allowed:\(C(n + r - 1,r) = \left( {\begin{array}{*{20}{c}}{n + r - 1}\\r\end{array}} \right) = \frac{{(n + r - 1)!}}{{r!(n - 1)!}}\)

with\(n! = n(n - 1) \cdot ..... \cdot 2 \cdot 1\)

Step 2(f): No broccoli croissants

The order of the elements does not matters (given), thus we need to use the definition of combination.

The shop has six types of croissants (plain, cherry, chocolate, almond, apple, broccoli) of which twenty-four (two dozen) have to be selected.

As we don’t need broccoli so only five croissants present. First we need to select one plain, two cherry, three chocolate, one almond and two apple croissant. Then we still need to select the remaining fifteen croissants from five remaining types of croissants (since broccoli croissants are no option)

We are interested in selecting \(r = 15\) elements from a set with \(n = 5\)elements.

Repetition of elements is allowed

\(\begin{array}{l}C(n + r - 1,r) = C(5 + 15 - 1,15)\\C(19,15) = \frac{\begin{array}{l}\\19!\end{array}}{{15!(19 - 15)!}} = \frac{{19!}}{{15!4!}} = 3876\end{array}\)

Step 3(f): One broccoli croissants

The order of the elements does not matters (given), thus we need to use the definition of combination.

The shop has six types of croissants (plain, cherry, chocolate, almond, apple, broccoli) of which twenty-four (two dozen) have to be selected.

First we need to select one plain, two cherry, three chocolate, one almond and two apple croissant. Then we still need to select the remaining fourteen croissants from five remaining types of croissants (since broccoli croissants are no option)

We are interested in selecting \(r = 14\) elements from a set with \(n = 5\)elements.

Repetition of elements is allowed

\(\begin{array}{l}C(n + r - 1,r) = C(5 + 14 - 1,14)\\C(18,14) = \frac{\begin{array}{l}\\18!\end{array}}{{14!(18 - 14)!}} = \frac{{18!}}{{14!4!}} = 3060\end{array}\)

Step 4(f): Two broccoli croissants

The order of the elements does not matters (given), thus we need to use the definition of combination.

The shop has six types of croissants (plain, cherry, chocolate, almond, apple, broccoli) of which twenty-four (two dozen) have to be selected.

As we don’t need broccoli so only five croissants present. First we need to select one plain, two cherry, three chocolate, one almond and two apple croissant. Then we still need to select the remaining thirteen croissants from five remaining types of croissants (since broccoli croissants are no option)

We are interested in selecting \(r = 13\) elements from a set with \(n = 5\)elements.

Repetition of elements is allowed

\(\begin{array}{l}C(n + r - 1,r) = C(5 + 13 - 1,13)\\C(17,13) = \frac{\begin{array}{l}\\17!\end{array}}{{13!(17 - 13)!}} = \frac{{17!}}{{13!4!}} = 2380\end{array}\)

Step 5(f): Three broccoli croissants

The order of the elements does not matters (given), thus we need to use the definition of combination.

The shop has six types of croissants (plain, cherry, chocolate, almond, apple, broccoli) of which twenty-four (two dozen) have to be selected.

As we don’t need broccoli so only five croissants present. First we need to select one plain, two cherry, three chocolate, one almond and two apple croissant. Then we still need to select the remaining twelve croissants from five remaining types of croissants (since broccoli croissants are no option)

We are interested in selecting \(r = 12\) elements from a set with \(n = 5\)elements.

Repetition of elements is allowed

\(\begin{array}{l}C(n + r - 1,r) = C(5 + 12 - 1,12)\\C(16,12) = \frac{\begin{array}{l}\\16!\end{array}}{{12!(16 - 12)!}} = \frac{{16!}}{{12!4!}} = 1820\end{array}\)

Step 6(f): No more than three broccoli croissants

The final answer can be found by using the Sum rule

From the results of step two three four and five we get\(3876 + 3060 + 2380 + 1820 = 11136\)

There are \(11136\) different ways.