Technique of Counting In a café , waiter had 6 orders from 6 different people but he forgot which order was for specific one of them(in other words he had orders in his hand but can not determine what each one of them had ordered) what is the number of ways such that 3 or more people can get their orders correctly ?
The answer I found was : 6C3 x 1 x (2 x1) + 6C4 x 1 x 1 + 0 + 1
I am conviced with 3rd and 4th terms , but I wounder why 6C3 was multiplied by 2 ?
 A: If you give A, B and C the correct items and D, E and  F the wrong items, the you can either 


*

*give E's item to D, F's item to E and D's item to F, or 

*give F's item to D, D's item to E and E's item to F.
This is related to derangements and rencontres numbers.
A: Add up the following:


*

*The number of ways such that exactly $3$ people get their orders correctly is $\binom63\cdot2=40$

*The number of ways such that exactly $4$ people get their orders correctly is $\binom64=15$

*The number of ways such that exactly $6$ people get their orders correctly is $\binom66=1$



Why multiply $\binom63$ by $2$?
Each of the other $3$ people must get someone else's order.
Let's mark these people $ABC$ and their orders $123$ respectively.
The possible permutations are:


*

*$\color{red}{A:1},\color{red}{B:2},\color{red}{C:3}$ "bad"  permutation

*$\color{red}{A:1},            B:3 ,            C:2 $ "bad"  permutation

*$            A:2 ,            B:1 ,\color{red}{C:3}$ "bad"  permutation

*$            A:2 ,            B:3 ,            C:1 $ "good" permutation

*$            A:3 ,            B:1 ,            C:2 $ "good" permutation

*$            A:3 ,\color{red}{B:2},            C:1 $ "bad"  permutation
As you can see, there are only $2$ cases where each one of them gets someone else's order.
