Question

There are 8 guests in a party. Each guest bring a gift and receives another gift in return . No one is allowed to receive the gift they bought. How many ways are there to distribute the gifts?​

Answer 1

Answer:

8!(1-(1/1!)+(1/2!)-(1/3!)+(1/4!)-(1/5!)+(1/6!)-(1/7!)+(1/8!))

that will be  14833.

Step-by-step explanation:

We can consider it as a dearrangement problem.

Since everyone will get 1 gift which is not the one they bring.

So that brings to the case of dearrangement.

Answer will be:

8!(1-(1/1!)+(1/2!)-(1/3!)+(1/4!)-(1/5!)+(1/6!)-(1/7!)+(1/8!))

that will be  14833.