Combinatoric factor in $\phi^4$-theory scattering I'm trying to work out the symmetry factor for a scattering inthe $\phi^4$ theory. 
My initial and final states are:
$\left|i\right> = \left|p_1,p_2 \right>$
$\left|f\right> = \left|p_3,p_4 \right>$
The expression for the amplitude to first order in $\lambda$ is of the form:
$$\left<f\right|\frac{-i\lambda}{4!}\phi(x)\phi(x)\phi(x)\phi(x)\left|i\right>$$
So then I have (ommiting the factors of energy which cancels out in the end anyway and the factor of $\frac{-i\lambda}{4!}$):
$$\left<0\right|\int d^4x \left[a_{p_3} a_{p_4} \int\int\frac{d^3q_1d^3q_2}{(2\pi)^6}a_{q_1}^\dagger a_{q_2}^\dagger e^{ix(q_1 + q_2)}\left|0\right>\left<0\right|\int\int\frac{d^3q_3d^3q_2}{(2\pi)^6}a_{q_3} a_{q_4} e^{-ix(q_3 + q_4)} a_{p_1}^\dagger a_{p_2}^\dagger\right] \left|0\right>$$
Now, using the commutation relations I commute the annihilation operators past the creation operators on the RHS and the other way around on the LHS. I drop some delta functions and I get:
$$\left<0\right|\int d^4x\left[\int\int\frac{d^3q_1d^3q_2}{(2\pi)^6}(2\pi)^6\left( \delta(p_3-q_1)\delta(p_4-q_2) + \delta(p_4-q_2)\delta(p_3-q_1)\right) e^{ix(q_1 + q_2)}\left|0\right>\left<0\right|\int\int\frac{d^3q_3d^3q_2}{(2\pi)^6}(2\pi)^6 \left( \delta(p_1-q_3)\delta(p_1-q_4) + \delta(p_1-q_4)\delta(p_2-q_3)\right)e^{-ix(q_3 + q_4)} \right]\left|0\right>$$
Which then gives me:
$$ \int d^4x (2e^{ix(p_3 + p_4)}) (2e^{-ix(p_1 + p_2)})  = 4\delta(p_3+p_4-p_1-p_2)$$
Which is what I expect except for the factor of 4. Because now bringing back the factor of $\frac{-i\lambda}{4!}$:
$$4\delta(p_3+p_4-p_1-p_2)\frac{-i\lambda}{4!}$$
But I know from other sources that the factor of $4!$ should get canceled. I think that by using different fields to annihilate different particles we get a combinatoric factor of $4!$ as we can arrange the four fields in this many possible permutations. But then the final result is by a factor of $4$ off. Why do I get this extra factor next to my $\delta$? 
Thanks!
PS. I couldn't find any info on how to break lines in long equations. Any tips for the future?
 A: See this answer How to count and 'see' the symmetry factor of Feynman diagrams? for more, I do not see any symmetry factors of $O(\lambda)$ in that diagram. The main reason is, is that there are no internal lines that can give you a symmetry factor on the $O(\lambda)$. 
Update 1:
Checking your reference the $4!$ comes from the $4!$ possible contractions because this is a scattering of identical particles so you can contract the fields $4!$ different ways that are identical so what you actually doing is summing up all possible contractions, that is not the same as symmetry factor which occurs on internal lines. 
Update 2 example:
So for example if we were to label the $\phi$'s even though they are identical say $\phi_1\phi_2\phi_3\phi_4$, then you can contract $\phi_1 $ with $p_1$, $\phi_2$ with $p_2$, $\phi_3 $ with $p_3$, and $\phi_4$ with $p_4$. But you could also have contracted  $\phi_1 $ with $p_2$, $\phi_2$ with $p_1$, $\phi_3$ with $p_3$, and $\phi_4$ with $p_4$ and so on ..., so since all these contractions are equal you can some over them and since there is $4!$ ways to contract you some over the $4!$ possibilities.
