Having some trouble with this combinatorics problem

[Calabi_Yau]

6 friends go to a party, each one carrying a different umbrella. They place the umbrellas outside. When the party is over, they are drunk and each one grabs an umbrella at random.
In how many ways could none of them have taken the right umbrella?

I'm having a bit trouble with this, as I can't seem to solve it without having to do some rough counting some times. Can any of you bother to solve this and explain it to me?

 

[Robert1986]

Take a look at this link: http://www.proofwiki.org/wiki/Hat-Check_Problem

 

[Norwegian]

You are asking about the number of derangments S_n of a set with n elements. If you are looking for an exact answer, you can either use the recurrence relation S_n+1=n(S_n+S_n-1), or compute the alternating sum Ʃ(-1)ⁱn!/i! where i goes from 0 to n.

 

[Calabi_Yau]

Thanks guys.

 

[CellCoree]

I can offer some insight into this combinatorics problem. First, we need to understand the concept of permutations and combinations. Permutations refer to the number of ways in which objects can be arranged in a specific order, while combinations refer to the number of ways in which objects can be selected without regard to order.

In this scenario, we are dealing with combinations, as the order in which the friends grab the umbrellas does not matter. We also need to consider that each friend can only grab one umbrella. Therefore, we are dealing with a combination without replacement.

To solve this problem, we can use the formula for combinations without replacement: nCr = n!/(r!(n-r)!), where n is the total number of objects (in this case, umbrellas) and r is the number of objects being selected (in this case, friends).

So, in this scenario, we have 6 friends (r) and 6 umbrellas (n). Therefore, the total number of ways in which they can grab the umbrellas is 6!/(6!(6-6)!) = 1. This means that there is only one way in which all 6 friends could have grabbed the wrong umbrellas.

I hope this explanation helps you understand the problem better. If you have any further questions, please feel free to ask.