Question

8. How many different ways are there to choose a dozen donuts from the 21 varieties at a donut shop?

Short answer

There are $225,729,840$different to choose a dozen donuts from the twenty-one varieties at a donut shop.

Step-by-step solution

Step 1: Definitions

Definition of Permutation (Order is important)

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

Repetition allowed:$n^T$

Definition of combination (order is important)

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

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

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

Step 2: Solution

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

We are interested in selecting r = 12 elements from a set with n = 21 elements.

Repetition of elements is allowed

$\begin{array}{r}C(n+r-1,r)=C(21+12-1,12)\\ C(7,3)=\frac{32!}{12!(32-12)!}=\frac{32!}{12!20!}=225,729,840\end{array}$