To solve this problem, we need to determine the probability that four randomly generated subsets of ({1, 2, 3, 4, 5}) have no elements in common and their union is ({1, 2, 3, 4, 5}).

**Total Subsets**: Each subset of ({1, 2, 3, 4, 5}) can be any of the (2^5 = 32) subsets. Since Tom presses the button 4 times, there are (32^4) possible outcomes.

**Valid Subsets**: We need to count the number of ways to choose four subsets such that:

Each element of ({1, 2, 3, 4, 5}) appears in exactly one of the four subsets.

No element is repeated in any subset.

**Distribution of Elements**: We can think of distributing the 5 elements into 4 distinct subsets. This is equivalent to finding the number of surjective (onto) functions from a set of 5 elements to a set of 4 elements.

**Counting Surjective Functions**: The number of surjective functions from a set of size (n) to a set of size (k) is given by:

k!⋅S(n,k)

where (S(n, k)) is the Stirling number of the second kind, representing the number of ways to partition (n) elements into (k) non-empty subsets.

For (n = 5) and (k = 4):

S(5,4)=10

Therefore, the number of surjective functions is:

4!⋅S(5,4)=24⋅10=240

**Probability Calculation**: The probability (P) that the four subsets have no elements in common and their union is ({1, 2, 3, 4, 5}) is:

P=324240​=1048576240​=6553615​

**Finding (p + q)**: The irreducible fraction is (\frac{15}{65536}), so (p = 65536) and (q = 15). Therefore, (p + q = 65536 + 15 = 65551).

So, (p + q) is **65551**.rm with this

The problem you’ve laid out involves determining the probability that four randomly generated subsets of the set \( S = \{1, 2, 3, 4, 5\} \) have no elements in common, and their union covers the entire set \( S \).

Let’s walk through the solution step by step:

### Step 1: Total Number of Possible Subsets
Each subset of \( S \) can be any subset of \( S \), so the total number of subsets of \( S \) is:
\[
2^5 = 32
\]
Since we are pressing the button 4 times, the total number of possible outcomes is:
\[
32^4 = 1048576
\]
This is the total number of ways to randomly select four subsets from \( S \), with each subset having 32 possibilities.

### Step 2: Counting Valid Subsets
We are tasked with ensuring that:
1. The union of the four subsets covers all the elements in \( S \).
2. No element is repeated in any subset (i.e., no element of \( S \) appears in more than one subset).

This problem can be viewed as distributing the 5 elements of \( S \) into 4 distinct subsets, with each element going to exactly one subset. This is equivalent to finding the number of surjective (onto) functions from a set of 5 elements to a set of 4 distinct subsets.

### Step 3: Surjective Function Calculation
The number of surjective functions from a set of size \( n \) to a set of size \( k \) is given by:
\[
k! \cdot S(n, k)
\]
where \( S(n, k) \) is the Stirling number of the second kind, which counts the number of ways to partition \( n \) elements into \( k \) non-empty subsets.

For \( n = 5 \) and \( k = 4 \), the Stirling number \( S(5, 4) \) is 10, meaning there are 10 ways to partition the set of 5 elements into 4 non-empty subsets. Thus, the number of surjective functions is:
\[
4! \cdot S(5, 4) = 24 \cdot 10 = 240
\]
This gives us the number of valid ways to assign each element of \( S \) to one of the four subsets such that no subset is empty and no element appears in more than one subset.

### Step 4: Probability Calculation
The probability \( P \) that the four randomly generated subsets have no elements in common and their union is \( S \) is:
\[
P = \frac{\text{Valid Subsets}}{\text{Total Subsets}} = \frac{240}{1048576} = \frac{15}{65536}
\]

### Step 5: Finding \( p + q \)
The irreducible fraction for the probability is \( \frac{15}{65536} \), where \( p = 65536 \) and \( q = 15 \). Therefore:
\[
p + q = 65536 + 15 = 65551
\]

Thus, the final answer is **65551**.

Would you like more details or have any questions about this solution?

### Further Questions:
1. How do Stirling numbers of the second kind relate to partitioning sets?
2. How can you generalize this problem for different set sizes?
3. What is the significance of surjective functions in combinatorics?
4. How does the probability change if we increase the number of subsets?
5. What are other examples of problems involving partitioning sets?

### Tip:
When dealing with set partitions and surjective functions, Stirling numbers of the second kind are a powerful tool for counting the ways to assign elements into distinct non-empty subsets.

To solve this problem, we need to determine the probability that four randomly generated subsets of ({1, 2, 3, 4, 5}) have no elements in common and their union is ({1, 2, 3, 4, 5}).

Explore the combinatorics and set theory behind determining the probability that four randomly generated subsets of {1, 2, 3, 4, 5} have no elements in common and their union covers the entire set. This problem uses surjective functions and Stirling numbers to calculate the result, suitable for undergraduate-level students.

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