For a positive integer $n$ let $t_n$ be the number of unordered triples of non-empty and pairwise disjoint subsets of a given set with $n$ elements. For example, $t_3 = 1$. Find a closed form formula for $t_n$ and determine the last digit of $t_{2022}$.

(I also give here that $t_4 = 10$, for a reader to check his/her understanding of the problem statement.)
1. **Counting Ordered Triples:**
   Each element in the set of \( n \) elements can be in one of the three subsets or in none of them. Therefore, there are \( 4^n \) ways to distribute the elements into three subsets (including the possibility of empty subsets).

2. **Excluding Cases with at Least One Empty Subset:**
   We need to exclude the cases where at least one of the subsets is empty. There are three choices for which subset is empty, and for each of these choices, the remaining elements can be in one of the other two subsets or in none of them. Thus, there are \( 3 \cdot 3^n \) such cases.

3. **Including Cases with at Least Two Empty Subsets:**
   We need to add back the cases where at least two of the subsets are empty, as these were subtracted twice in the previous step. There are three choices for which two subsets are empty, and the remaining elements can either be in the third subset or not. Thus, there are \( 3 \cdot 2^n \) such cases.

4. **Excluding the Case Where All Subsets Are Empty:**
   Finally, we need to exclude the unique case where all three subsets are empty, which we have added back in the previous step. There is exactly 1 such case.

5. **Combining All Cases:**
   Using the principle of inclusion-exclusion, the total number of ordered triples is:
   \[
   4^n - 3 \cdot 3^n + 3 \cdot 2^n - 1
   \]

6. **Counting Unordered Triples:**
   Since the triples are unordered, we need to divide by the number of permutations of three items, which is \( 3! = 6 \). Therefore, the number of unordered triples is:
   \[
   t_n = \frac{4^n - 3 \cdot 3^n + 3 \cdot 2^n - 1}{6}
   \]

7. **Finding the Last Digit of \( t_{2022} \):**
   To find the last digit of \( t_{2022} \), we need to compute \( t_{2022} \mod 10 \). We start by finding the last digits of \( 4^{2022} \), \( 3^{2023} \), and \( 2^{2022} \).

   - The last digit of \( 4^n \) cycles every 2: \( 4, 6 \). Since \( 2022 \) is even, the last digit of \( 4^{2022} \) is \( 6 \).
   - The last digit of \( 3^n \) cycles every 4: \( 3, 9, 7, 1 \). Since \( 2023 \mod 4 = 3 \), the last digit of \( 3^{2023} \) is \( 7 \).
   - The last digit of \( 2^n \) cycles every 4: \( 2, 4, 8, 6 \). Since \( 2022 \mod 4 = 2 \), the last digit of \( 2^{2022} \) is \( 4 \).

   Now, we compute:
   \[
   4^{2022} \mod 10 = 6
   \]
   \[
   3 \cdot 3^{2023} \mod 10 = 3 \cdot 7 = 21 \mod 10 = 1
   \]
   \[
   3 \cdot 2^{2022} \mod 10 = 3 \cdot 4 = 12 \mod 10 = 2
   \]
   \[
   4^{2022} - 3 \cdot 3^{2023} + 3 \cdot 2^{2022} - 1 \mod 10 = 6 - 1 + 2 - 1 = 6
   \]
   Finally, we divide by 6 and find the last digit:
   \[
   t_{2022} \mod 10 = \frac{6}{6} = 1
   \]

The final answer is \( \boxed{ 1 } \)