\section*{Exercise 3 - 011213}

How many different three-digit numbers can be formed using the digits
a) 1 and 2,
b) 1, 2, and 3,
c) 1, 2, 3, and 4,

where the digits can be used multiple times? Try to find a pattern!

1) What solution do you get for four-digit numbers?

2) What can be conjectured for four-digit numbers if \(n\) digits are available? Try to prove this conjecture!
}

We are dealing with variations with repetition here, as we want to select \(k\), not necessarily distinct elements from the set of the first \(n\) natural numbers and write them in a sequence (thus considering the order). For each of the \(k\) positions in the sequence, we have \(n\) possibilities, so the number sought \(V(n, k)\) is given by

\[
V(n, k)=\underbrace{n \cdot n \cdot \ldots \cdot n}_{\text{k times}}=n^{k}
\]

Thus, we can form

a) \(V(2,3)=2^{3}=8\)

b) \(V(3,3)=3^{3}=27\)

c) \(V(4,3)=4^{3}=64\) different three-digit numbers.

d) For four-digit numbers, we find analogously \(2^{4}=16, 3^{4}=81 \text{ or } 4^{4}=256\) different solutions.

e) If, on the other hand, we have \(n\) digits available, then
- \(n\) numbers with four identical digits aaaa can be specified,
- \(4 n(n-1)\) numbers with three identical digits aaab can be specified (\(n\) possibilities to choose the digit a, \(n-1\) possibilities to choose the digit b, and 4 positions where b can stand),
- \(3 n(n-1)\) numbers of the form aabb can be specified (\(n\) possibilities to choose the digit a, \(n-1\) possibilities to choose the digit b, and 3 possibilities for the placement of the two pairs aa and bb),
- \(6 n(n-1)(n-2)\) numbers of the form aabc can be specified (\(n\) possibilities to choose the digit a, \(n-1\) possibilities to choose the digit b, \(n-2\) possibilities to choose the digit c, and \(\binom{4}{2}=6\) possibilities for the placement of the digits b and c or the digits a and a),
- finally, \(n(n-1)(n-2)(n-3)\) numbers abcd with four different digits can be specified.

The total number is thus

\[
n+4 n(n-1)+3 n(n-1)+6 n(n-1)(n-2)+n(n-1)(n-2)(n-3)=n^{4}
\]