combinatorics - family question We have 6 grandchildren.  2 from each of our 3 children.  Our granddaughter asked the following question."How many ways could they draw names for Christmas if they could not draw their sibling."  I have the answer, 80, but I would like to know the mathematical algorithm to get this answer. 
 A: Order the pairs of siblings like this $\underbrace{**}_\text{siblings}|\underbrace{**}_\text{siblings}|\underbrace{**}_\text{siblings}$
Then what you want is the number of permutations such that none of them are in the same space between bars as before. 
There are three cases. 
Case 1: siblings 1 and 2 fall into positions 3 and 4 (2 orders). So then siblings 3 and 4 fall in positions 5 and 6 (2 orders).siblings 5 and 6 fall into positions 1 and 2(2 ways) 8 total.
Case 2: one of siblings 1 and 2 falls into one of positions 3 and 4 and the other one falls into positions 5 and 6. (4 ways to chose exact positions where they fall and 2 to chose where each one falls gives 8 choices).This leaves the siblings 3 and 4 with three acceptable positions, however they need to take the vacant spot in positions 5,6 or the sibling 5,6 won't have enough space. (so they have 2 choices to pick the spot in spaces 1,2 and 2 choices to pick which sibling gets which gives 4 choices for siblings 3,4. Finally siblings 5,6 have two choices to see which one gets which of the 2 remaining spots. all in all 64 choices.
Case 3: siblings 1 and 2 pick spots 5 and 6 (2 orders). So siblings 3 and 4 pick 1 and 2 (2 orders) and siblings 5 and 5 pick spots 3,4 (2 orders) 8 total.
Therefore there are 8+64+8=80 choices. 
A: The set $G:=\{a,b,c\}\times\{1,2\}$  of grandchildren is the union of three couples $\hat x:=\{(x,1), (x,2)\}$ with $x\in \{a,b,c\}$. We are told to count the number of bijections $f:\>G\to G$ that satisfy the condition $$f_1(x,\iota)\ne x\qquad \forall\ (x,\iota)\in G\ ,$$ which expresses that no gift should go to oneself or to ones sibling. Call these $f$ admissible. 
Let $f$ be admissible. The couple $\hat x$ is mixing when $f_1(x,1)\ne f_1(x,2)$. When one couple is mixing all have to be: Assume that $\hat a$ is mixing and $\hat b$ is not. Then necessarily $f(\hat b)=\hat a$. It would then be impossible to define $f\restriction\hat c$ in an admissible way.
Therefore there are two types of admissible $f$'s, called mixing and factoring. The types of $f$ and $g:=f^{-1}$ are the same.
A mixing $f$ can be specified in the following way: For each couple $\hat x$ admissible values $f_1(x,1)$ and $g_1(x,1)$ are chosen independently. This then determines admissible values $f_1(x,2)$ and $g_1(x,2)$, and it is easy to see that $f$ is uniquely specified by these data. There are $6$ independent binary choices involved; so there are $64$ mixing admissible $f$'s.
A factoring $f$ acts in the way $\hat a\to\hat b\to\hat c\to\hat a$ or vice versa. At each arrow one binary choice is possible, giving a total of $16$ possibilities.
It follows that there are $80$ admissible bijections. The probability that a random drawing results in an admissible $f$ comes to  ${1\over 9}$ only.
