OPE double-contractions between $T$ and $e^{ikX}$ I am reading David Tong's lecture notes chapter 4 http://www.damtp.cam.ac.uk/user/tong/string.html
On the top of page 82 in the eq. before eq. (4.27), we are computing the OPE between $T$ and $e^{ikX}$ using Wick's Theorem, it says

I wonder why the first term does not have an extra coefficient of 2?
Since there are two $\partial X$ in the energy momentum tensor $T$, isn't there two ways of doing 2-contractions of $T$ and $e^{ikX}$, just like the second term?
So why doesn't the first term have a 2 like the second term?
 A: TL;DR: No, the order of performing the double-contraction$^1$ should be moded out, i.e. one should only count the number of pairs of double-contractions. 
Tip: To not make combinatoric mistakes, one might want to consider a monomial first instead of the full exponential/vertex operator. For instance$^2$ 
$${\cal R}( :X(z)^n::X(w)^m:)$$ leads to 
$$ n  (n-1) \times m  (m-1) \times \frac{1}{2!} \text{  double contractions}, $$
and so forth. 
--
$^1$ Note to the reader: In Tong's normalization, each contraction comes with a factor $\frac{\alpha^{\prime}}{2}$, cf. 2nd last formula on p. 80. 
$^2$ Here ${\cal R}$ denotes the often implicitly written radial ordering. The OPE calculation consists of evaluating a nested Wick's theorem between radial ordering ${\cal R}$ and normal ordering $::$, cf. my Phys.SE answer here. 
A: The contractions are given by:
$$
:A::B:=\exp{\int -\frac{\alpha'}{2}\eta^{\mu\nu}\ln |z_1-z_2|^2\,\delta_{X^{\mu}(z_1,\bar{z}_1)}^{(A)}\delta^{(B)}_{X^{\nu}(z_2,\bar{z}_2)}} :AB:
$$
In the case where $B=e^{ikX(z,z)}$ note that $B$ is an eigenfunctional of $\delta^{B}_{X^{\nu}(z_2,\bar{z}_2)}$:
$$
\delta^{B}_{X^{\nu}(z_2,\bar{z}_2)}e^{ikX(z,z)}=ik_{\nu}\delta^2(z-z_2)e^{ikX(z,\bar{z})}
$$
so we just need to do $\delta^{B}_{X^{\nu}(z_2,\bar{z}_2)}\rightarrow ik_{\nu}\delta^2(z-z_2)$, getting
$$
:A::e^{ikX(z,\bar{z})}:=\exp{\int -\frac{\alpha'}{2}\eta^{\mu\nu}\ln |z_1-z_2|^2\,\delta_{X^{\mu}(z_1,\bar{z}_1)}^{(A)}(ik_{\nu}\delta^2(z-z_2)}) :Ae^{ikX(z,\bar{z})}:
$$
and then the contraction will be just in $A$, doing 
$$
X(z',\bar{z}')^{\mu}\rightarrow \frac{\alpha'}{2}ik^{\mu}\ln|z'-z|^2
$$
In your case where $A=\partial X(z,\bar{z}).\partial X(z,\bar{z})$ we have $2$ ways of contracting just one of the $X$'s and only one way of contracting both the $X$'s
