The five oceans

It's just derangement of $5$ which equals $44$ .

For details about derangements, check out this wiki from Brilliant: Derangements

In an arrangement, when no element goes to its specified position or all elements go to wrong positions , it's called Derangements .

Let $D(n)$ be the number of derangements for $n$ different objects, then $D(n)=n!\displaystyle\sum_{r=0}^{n} \dfrac{(-1)^r}{r!}.$

Another form of derangement is $D(n)=n!\left(\frac { 1 }{ 2! } -\frac { 1 }{ 3! } +\frac { 1 }{ 4! }-\cdots+{ (-1) }^{ n }\frac { 1 }{ n! } \right).$

Then, the number of derangements for $5$ different objects is $D(5)=5!\left(\frac { 1 }{ 2! } -\frac { 1 }{ 3! } +\frac { 1 }{ 4! } -\frac { 1 }{ 5! }\right).$ $=\boxed{44}.$

Putting the labels on the oceans is a permutation of 5 objects, so there are 5! ways of doing this. Permutations can be written as a product of cycles. For example (1,2,3)(4,5) means 1 -> 2 -> 3 ->1 and 4 -> 5 -> 4; while (1,2,3,4)(5) means 1 -> 2 -> 3 -> 4 -> 1 and 5 -> 5 (unchanged). Usually, cycles of length 1 are not listed explicitly. Now, for all the labels to be incorrect, we must have a permutation with no cycles of length 1 (which is an element that is not moved). For five objects, that leaves permutations of the forms (1,2,3,4,5) and (1,2,3)(4,5). In each case, there are 5! ways of assigning numbers to the cycles, but cycles of length n will occur n times since (1,2,3), (2,3,1), and (3,1,2) are the same permutations. Thus there are 5!/5 = 24 permutations with a single 5-cycle. and 5!/(2x3) = 20 permutations of the form (1,2,3)(4,5). So there are 44 permutations that are completely incorrect.

Step 1. The number of ways in which all labels are wrong = Total number of ways to arrange labels - Number of ways at least 1 label is correct.

Step 2. Total number of label arrangements: 5!

Step 3. Use PIE to calculate N, the number of way at least 1 label is correct:

Number of ways ocean A is labeled correctly: 1 * 4!

There are 5 oceans

-> 5 * 4! = 5! ways for all the ways to get 1 ocean right.

Number of ways both ocean A and ocean B are labeled correctly: 1 * 1 * 3!

There are 5 * 4 / 2! ways to select oceans A and B

-> $\frac{3! * 5 * 4}{2!}$ = $\frac{5!}{2!}$ ways for all the ways to get 2 oceans right.

Number of ways ocean A, B, C are labeled correctly: 1 * 1 * 1 * 2!

There are $\frac{5 * 4 * 3}{3!}$ 5 * 4 * 3 / 3! ways to select oceans A, B, C

-> $\frac{5 * 4 * 3 * 2!}{3!}$ = $\frac{5!}{3!}$ ways for all the ways to get 3 oceans right.

Similarly, $\frac{5!}{4!}$ for all the ways to get 4 oceans right, and $\frac{5!}{5!}$ for all the ways to get 5 oceans right.

Using PIE, N = 5! - $\frac{5!}{2!}$ + $\frac{5!}{3!}$ - $\frac{5!}{4!}$ + $\frac{5!}{5!}$

Step 4. Our answer is:

5! - N = 5! - 5! + $\frac{5!}{2!}$ - $\frac{5!}{3!}$ + $\frac{5!}{4!}$ - $\frac{5!}{5!}$ = 5! (1 - $\frac{1!}{1!}$ + $\frac{1!}{2!}$ - $\frac{1!}{3!}$ + $\frac{1!}{4!}$ - $\frac{1!}{5!}$ )

       = 44

A000166

2 1 4 5 3 ; 3 1 2 5 4 ; 4 1 2 5 3 ; 5 1 2 3 4

2 1 5 3 4 ; 3 1 4 5 2 ; 4 1 5 2 3 ; 5 1 4 2 3

2 3 1 5 4 ; 3 1 5 2 4 ; 4 1 5 3 2 ; 5 1 4 3 2

2 3 4 5 1 ; 3 4 1 5 2 ; 4 3 1 5 2 ; 5 3 1 2 4

2 3 5 1 4 ; 3 4 2 5 1 ; 4 3 2 5 1 ; 5 3 2 1 4

2 4 1 5 3 ; 3 4 5 1 2 ; 4 3 5 1 2 ; 5 3 4 1 2

2 4 5 1 3 ; 3 4 5 2 1 ; 4 3 5 2 1 ; 5 3 4 2 1

2 4 5 3 1 ; 3 5 1 2 4 ; 4 5 1 2 3 ; 5 4 1 2 3

2 5 1 3 4 ; 3 5 2 1 4 ; 4 5 1 3 2 ; 5 4 1 3 2

2 5 4 1 3 ; 3 5 4 1 2 ; 4 5 2 1 3 ; 5 4 2 1 3

2 5 4 3 1 ; 3 5 4 2 1 ; 4 5 2 3 1 ; 5 4 2 3 1

{0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961, 14684570, 176214841, 2290792932, 32071101049, 481066515734, 7697064251745, 130850092279664, ...} formed a series.

The 11 x 4 = 44 {Answer}