Number of ways in which n distinct things can be distributed among n p...
There are n! ways in which everyone gets a thing. There are n^n distributions in total.
Hence the formula for what you want is n^n−n!.
We start trying with n = 4 and get 4^4 − 4! = 232.
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Number of ways in which n distinct things can be distributed among n p...
To solve this problem, we need to find the value of 'n' for which the number of ways in which n distinct things can be distributed among n persons so that at least one person does not get anything is 232.
Let's break the problem down step by step:
1. Total number of ways to distribute 'n' distinct things among 'n' persons:
The total number of ways to distribute 'n' distinct things among 'n' persons can be calculated using the concept of permutations. Each person can be assigned one of the 'n' distinct things, and there are 'n' options available for the first person, 'n-1' options available for the second person, and so on. Therefore, the total number of ways is given by n!
2. Number of ways in which every person gets at least one thing:
To find the number of ways in which every person gets at least one thing, we can use the concept of derangements. A derangement is a permutation of a set of objects in which none of the objects appear in their original position. The number of derangements of 'n' objects is given by the formula:
D(n) = n!(1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n/n!)
3. Number of ways in which at least one person does not get anything:
The number of ways in which at least one person does not get anything can be calculated by subtracting the number of ways in which every person gets at least one thing from the total number of ways to distribute 'n' things among 'n' persons.
Number of ways = n! - D(n)
4. Given that the number of ways is 232:
n! - D(n) = 232
To solve this equation, we can substitute different values of 'n' and calculate the corresponding values of D(n) until we find a solution.
Let's calculate the values of D(n) for different values of 'n':
For n = 1:
D(1) = 1!(1 - 1/1!) = 0
1! - D(1) = 1 - 0 = 1 (not equal to 232)
For n = 2:
D(2) = 2!(1 - 1/1! + 1/2!) = 2(1 - 1 + 1/2) = 1
2! - D(2) = 2 - 1 = 1 (not equal to 232)
For n = 3:
D(3) = 3!(1 - 1/1! + 1/2! - 1/3!) = 6(1 - 1 + 1/2 - 1/6) = 2
3! - D(3) = 6 - 2 = 4 (not equal to 232)
For n = 4:
D(4) = 4!(1 - 1/1! + 1/2! - 1/3! + 1/4!) = 24(1 - 1 + 1/2 - 1/6 + 1/24) = 9
4! - D(4) = 24 - 9 = 15 (