a Miranda bought a new bicycle lock with a four-number combination in which the numbers are from $0$ to $9.$ We will find the number of possible combinations when there are no restrictions on the repetitive usages of numbers. To find it we will use the Fundamental Counting Principle. Let's list the number of possible outcomes of each event.
$\stackrel{\normalsize{\textbf{Number of Choices}}}{\textbf{ for the First Number}}$
$\col{10}$
$\stackrel{\normalsize{\textbf{Number of Choices}}}{\textbf{ for the Second Number}}$
$\colII{10}$
$\stackrel{\normalsize{\textbf{Number of Choices}}}{\textbf{ for the Third Number}}$
$\colIV{10}$
$\stackrel{\normalsize{\textbf{Number of Choices}}}{\textbf{ for the Fourth Number}}$
$\colV{10}$
Notice that since the number choices have no restrictions and there are $10$ numbers from $0$ to $9,$ each event has $10$ possible choices. Now we will multiply the number of choices for each event to find the number of possible combinations.
\begin{gathered}
\col{10} \times \colII{10} \times \colIV{10} \times \colV{10} = 10\, 000
\end{gathered}
There are $10\, 000$ possible combinations of numbers for determining the lock.
b We will find the number of possible combinations when Miranda can use each number for the lock only once. Therefore, after determining a number for the first event we cannot use this number for other events, which means that after each event we will reduce the number of choices by one.
$\stackrel{\normalsize{\textbf{Number of Choices}}}{\textbf{ for the First Number}}$
$\col{10}$
$\stackrel{\normalsize{\textbf{Number of Choices}}}{\textbf{ for the Second Number}}$
$\colII{9}$
$\stackrel{\normalsize{\textbf{Number of Choices}}}{\textbf{ for the Third Number}}$
$\colIV{8}$
$\stackrel{\normalsize{\textbf{Number of Choices}}}{\textbf{ for the Fourth Number}}$
$\colV{7}$
Now we will use the Fundamental Counting Principle to find the desired number of possible combinations.
\begin{gathered}
\col{10} \times \colII{9} \times \colIV{8} \times \colV{7} = 5040
\end{gathered}
Miranda has $5\040$ different number combinations to determine the lock with the given restriction.
