$\DeclareMathOperator\RT{RT}$I am reading the book *"Algebraic operads"* by J. L. Loday and B. Vallete. The authors have given a combinatorial construction of the free operad over an $\mathbb{S}$-module $M$ in Section 5.5, pg. 126. I have a question regarding this construction, which I am describing below.

Let $\RT(n)$ denote the set of all nonplanar rooted trees with $n$ leaves such that each vertex has at least one input. For any tree $t \in \RT(n)$, let $\mathrm{vert}(t)$ be the set of all vertices of $t$ and $\mathrm{in}(v)$ denote the number of inputs of the vertex $v \in \mathrm{vert}(t)$.

For any $\mathbb{S}$-module $M=(M(0),M(1),M(2),\ldots)$ with $M(0)=0$ and $t \in \RT(n)$, the *treewise tensor product* of $M$, denoted $M(t)$, is defined as
$$M(t) := \bigotimes_{v ~\in~ \mathrm{vert}(t)} M\big(in(v)\big)$$

**The free operad $\mathbb{T}(M)$ over $M$:**

The free operad $\mathbb{T}(M)= (\mathbb{T}(M)(0), \mathbb{T}(M)(1), \mathbb{T}(M)(2) ,\ldots)$ over the $\mathbb{S}$-module $M$ is defined as $$\mathbb{T}(M)(n) := \bigoplus_{t ~\in ~\RT(n)} M(t)$$ for all $n \ge 0$.

The authors in Section 7.6, p. 193 mentions that for a given binary quadratic operad $\overline{E} = (0,0,E,0,\ldots)$ the lower degree modules of the free operad $T(\overline{E})$ is described as $$T(\overline{E})(0) = 0, \quad T(\overline{E})(1) = \mathbb{K}, \quad T(\overline{E})(2) = E, \quad T(\overline{E})(3) = 3 (E \otimes E)$$

**My question:**

I am convinced by the fact that $T(\overline{E})(2) = E$. But how are the authors arriving at the fact that $T(\overline{E})(3) = 3(E \otimes E)?$ According to the definition mentioned above of the free operad we get that
$$T(\overline{E})(3) = \bigoplus_{t \in \RT(3)} \overline{E}(t)$$

We know that $\RT(3)$ contains only three elements as shown below:

Then we get $T(\overline{E})(3) = \overline{E}(2) \otimes \overline{E}(2) \bigoplus \overline{E}(2) \otimes \overline{E}(2) \bigoplus \overline{E}(3) = 2 (E \otimes E)$. But how do the authors get $T(\overline{E})(3) = 3(E \otimes E)$?

**Edit 1:**

Following Todd Trimble's answer if we try to give an explicit description of $T(\overline{E})(4)$. Then we must find the collection of all subsets of $\{1,2,3,4\}$ satisfying the relations mentioned in Todd Trimble's answer which gives us $19$ possibilities:
$$\big\{\{1,2,3\}, \{1,2,3,4\}\big\}, \big\{\{1,2\}, \{1,2,3\}, \{1,2,3,4\}\big\}, \big\{\{1,3\}, \{1,2,3\}, \{1,2,3,4\}\big\}, \big\{\{2,3\}, \{1,2,3\}, \{1,2,3,4\}\big\}$$

$$\big\{\{1,2,4\}, \{1,2,3,4\}\big\}, \big\{\{1,2\}, \{1,2,4\}, \{1,2,3,4\}\big\}, \big\{\{1,4\}, \{1,2,4\}, \{1,2,3,4\}\big\}, \big\{\{2,4\}, \{1,2,4\}, \{1,2,3,4\}\big\}$$

$$\big\{\{1,3,4\}, \{1,2,3,4\}\big\}, \big\{\{1,3\}, \{1,3,4\}, \{1,2,3,4\}\big\}, \big\{\{1,4\}, \{1,3,4\}, \{1,2,3,4\}\big\}, \big\{\{3,4\}, \{1,3,4\}, \{1,2,3,4\}\big\}$$

$$\big\{\{2,3,4\}, \{1,2,3,4\}\big\}, \big\{\{2,3\}, \{2,3,4\}, \{1,2,3,4\}\big\}, \big\{\{2,4\}, \{2,3,4\}, \{1,2,3,4\}\big\}, \big\{\{3,4\}, \{2,3,4\}, \{1,2,3,4\}\big\}$$

$$\big\{\{1,2\},\{3,4\},\{1,2,3,4\}\big\}, \big\{\{1,3\},\{2,4\},\{1,2,3,4\}\big\}, \big\{\{1,4\},\{2,3\},\{1,2,3,4\}\big\}$$

The possibilities $\big\{\{1,2,3\}, \{1,2,3,4\}\big\}, \big\{\{1,2,4\}, \{1,2,3,4\}\big\}, \big\{\{1,3,4\}, \{1,2,3,4\}\big\}$, and $ \big\{\{2,3,4\}, \{1,2,3,4\}\big\}$ indexes a trivial summand (because $\overline{E}(3)=0$). Therefore, that leaves us with the remaining fifteen, where the treewise tensor products are each isomorphic to $E \otimes E \otimes E$.

Therefore, $T(\overline{E})(4) = 15 (E \otimes E \otimes E)$. Is this correct?

**Edit 2:**

Following Todd Trimble's answer, I have a small question regarding the construction of the free operad. The set $\{\{1,3\},\{1,2,3\}\}$ corresponds to the following graph

is this correct? If yes, then what about this graph

will this not be present in the $\mathbb{T}(E)(3)?$ It looks like this graph also corresponds to the set $\{\{1,3\},\{1,2,3\}\}$.