The Probability of a Pillow Case Returning to Its Room After Laundry: A Mathematical Analysis
Imagine you are a guest at a hotel that is fully booked, and you need to get your pillow case laundered. With all the rooms occupied, the chances of your pillow case ending up back in your room after the laundry process are quite intriguing.
Given the scenario where a hotel has n rooms, and each room has a pillow, what are the odds that a specific pillow case will return to the same room it came from?
Analyzing the Probability
The problem can be approached through permutations and combinatorial analysis. Let's break down the possible scenarios:
Scenario 1: One Pillow Returns Correctly
Consider the case where exactly one pillow returns to its original room. This means that the remaining n-1 pillows are permuted such that none of them return to their original rooms. The number of ways to do this is given by:
(n-1)!-1 permutations where none of the pillows return to their original rooms (Derangement formula).Since there are n rooms, there are n such cases where exactly one pillow returns to its original room. Therefore, the total number of ways for this scenario is:
n * ( (n-1)! - 1 )
Scenario 2: Multiple Pillows Return Correctly
Let's consider the case where exactly two pillows return to their original rooms. This means the remaining n-2 pillows are permuted such that none of them return to their original rooms. The number of ways to do this is:
(n-2)!-1 permutations for the remaining pillows.The number of ways to choose 2 pillows out of n to return correctly is given by:
C(n,2) which is (n * (n-1)) / 2.
Therefore, the total number of ways for this scenario is:
C(n,2) * ( (n-2)! - 1 )
Generalizing to k Pillows Returning Correctly
For the general case where exactly k pillows return to their original rooms, the remaining n-k pillows are permuted such that none of them return to their original rooms. The number of ways to choose k pillows out of n is:
C(n,k) which is n! / (k! * (n-k)!).
The number of ways to permute the remaining n-k pillows is:
(n-k)!-1
Therefore, the total number of ways for this scenario is:
C(n,k) * ( (n-k)!-1 )
Summing up all these scenarios from k1 to kn-2, and including the case where all pillows return to their original rooms (which is 1 way), we get:
Sum from k1 to n-2 of [ C(n,k) * ( (n-k)!-1 ) ] 1
Calculation of Total Permutations
The total number of permutations of n items is:
n!
Therefore, the probability that a specific pillow case will return to its original room is:
Sum from k1 to n-2 of [ C(n,k) * ( (n-k)!-1 ) ] 1 out of n!
Conclusion
The analysis above provides a detailed mathematical framework to understand the probability of a pillow case returning to its original room after being laundered in a fully booked hotel. The odds can be calculated using the summation of derangements and combinations, reflecting the complex distribution of items in a constrained environment.