Tins of paint/lids

[Garty22]

You have 6 (or any number) of tins of different coloured paint. Each one has its own correct lid. All the lids have been taken off. If you close your eyes and put each lid back on to a random tin, what are the chances of at least one of lids being on the correct paint tin?

 

[Harry_the_cat]

Use the complement. Find the prob that none of the paint lids are correct and subtract from 1.

 

[pka]

You have 6 (or any number) of tins of different coloured paint. Each one has its own correct lid. All the lids have been taken off. If you close your eyes and put each lid back on to a random tin, what are the chances of at least one of lids being on the correct paint tin?

This is the well known idea of a derangement; the idea of rearranging \(N\) items so that no one is in its original place.
Look at the provided link. Scroll down to the table. See there \(\mathcal{D}(6)=265\).
Consider the string \(ABCDEF\), six letters, there are \(6!=720\) possible rearrangements, \(265\) of which have no letter in its correct position.
This expression is useful \(\mathcal{D}(n)=\left\lfloor {\dfrac{{n!}}{e}}+\dfrac{1}{2} \right\rfloor \)

 

[Steven G]

I recently studied about derangement and am just amazed hoe the number keeps popping up seemingly out of nowhere!