# Thread: Tricky Permutation/Graph theory problem

1. ## Tricky Permutation/Graph theory problem

Count the permutations θ in $S_6$ that satisfy θ(1)=2 and θ(2)=3.

I'm mostly confused by the wording of the problem. Would it be implying that a permutation can ONLY contain those criteria OR that it can contain those two and any other subsequent "run".

2. Originally Posted by elninio
Count the permutations θ in $S_6$ that satisfy θ(1)=2 and θ(2)=3.
If I were given this problem as written, I would answer $4!$.
That is the number of ways to assign $3,4,5,~\&~6$.

3. My understanding is that the permutations do no have to include 3,4,5 and 6.

4. Originally Posted by elninio
My understanding is that the permutations do no have to include 3,4,5 and 6.
This may well be a case where notation not being standard is the difficulty.
The symbol $S_6$ usually stands for the symmetric group on six elements.
It is a group of order $6!=720$ with group operation permutation composition.
Here is an example: $\theta =\left( {\begin{array}{cccccc}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 3 & 1 & 4 & 6 & 5 \\ \end{array} } \right)$
.

So in this is standard notation $3,4,5,~\&~6$ must be assigned.

Does this differ with the definition in your textbook?

5. It seems to be the second case, where there could be a different amount of members that satisfy theta.
I'm confused as to how I would account for the possibility of it including 3,4,5 and all 6 members.

There are $4!$ permutations in $S_6$ each of which assigns $1\to 2~\&~2\to 3$.