(Iso)spin coupling of three particles I need Isospin calculations to predict the relative amount of strong decays, namely $D^{**0}\to D\pi\pi$, where the $D$ and $\pi$ can be charged or neutral.
My problem is the order of contracting the states, here an example:
$D^{**0}\to D^0\pi^0\pi^0$, represented by $\left\langle \frac{1}{2},-\frac{1}{2}\middle|\frac{1}{2}, -\frac{1}{2}\right\rangle|1, 0\rangle|1, 0\rangle$
In principle I know how to do this with all the Clebsh-Gordan Coefficient.
However I get different results if I first contract the $D$ with one $\pi$, compared to contracting both pions fist:
$(\left|\frac{1}{2}, -\frac{1}{2}\right\rangle|1, 0\rangle)|1, 0\rangle \neq \left|\frac{1}{2}, -\frac{1}{2}\right\rangle(|1, 0\rangle|1, 0\rangle)$
What did I miss there?
Thank you in advance!
 A: OK, I'll get to the answer with insouciance that would make Paul Gordan turn in his grave, and only then, separately, I'll dismiss your dilemma by summary reassurance via the Racah recouping coefficients. This is not the most complete answer, but it is by far the fastest one, exploiting a non-universal piece of luck, here. I'll be pretending isospin is spin here, without loss of generality, so then $j_1=1/2=-m_1; ~ j_2=j_3=1; ~ m_2=m_3=0$. 
The states $|a\rangle \equiv |j_1,m_1\rangle ~|j_2,m_2\rangle ~|j_3,m_3\rangle$ are a small subset of the 2x3x3=18 eigenstates of the 6 commuting operators $j_1,m_1,j_2,m_2,j_3,m_3$. Your left hand side path, $\equiv |b\rangle$, is in the basis of eigenstates of the slightly different 6 operators 
$J,M,J_{12},M_{12},j_3,m_3$. Your right hand side is eigenstates of the 6 yet different operators $J,M,j_1,m_1,J_{23},M_{23}$. 
These three bases are equivalent:  they are linearly transformable to each other by Clebsch-Gordan coefficients and /or Wigner 6j symbols, or, better yet, equivalent Racah recouping coefficients W. 
Dumb luck: since 2 and 3 are symmetric, only $J_{23}=0$ contributes to $\langle 1/2, -1/2~|~c\rangle$, the symmetry being reflected in the Clebsch-Gordan coefficients, so 
$$
|c\rangle= \sqrt{2/5}~|~5/2,-1/2 \rangle+ \sqrt{4/15}~|~3/2,-1/2 \rangle  -  1/\sqrt{3}~|~ 1/2,-1/2\rangle. 
$$
So $\langle 1/2,-1/2~|~c\rangle=-1/\sqrt{3}$, and 1/3 in the rate, through your right hand side channel, Kronecker-combining the pions first.
Now, for philosophical reassurance. Your left hand side calculation is messier, since now $\langle 1/2,-1/2~|~b\rangle$ gets support from the $J_{12}=3/2$ channel,$\sqrt{2/3}~|~J_{12}=3/2,M_{12}=-1/2\rangle ~|1,0\rangle$ , as well as the  $J_{12}=1/2$ channel, $\sqrt{1/3}~|~J_{12}=1/2,M_{12}=-1/2\rangle ~|1,0\rangle$ . One does not just add them unweighted, indiscriminately. Since $J_{12}$ and $J_{23}$ do not commute, particle 2 (pion) being entangled differently, $|~b\rangle \neq |~c\rangle$. Their bases need to be rotated to each other by the Racah coefficients $\langle ...|~...\rangle$ below.
For $J=1/2, M=-1/2$, skipping them here to uncluttered the (somewhat paradoxical looking connection coefficients without them), we have the change of basis $ \langle b |c\rangle$,
$$
|j_1,J_{23}\rangle= \sum_{J_{12}} |J_{12}, j_3\rangle\langle J_{12},j_3|j_1, J_{23}\rangle,
$$
where the needed Racah coeffs are $\langle {3/2},1|1/2,0\rangle =\sqrt{2/3} $  and $\langle {1/2},1|1/2,0\rangle =\sqrt{1/3} $.
So, upon this basis change, the left hand side would have "simplified" to the straightforward right hand side. 
To be sure, this is the lazy man's answer, but when one sees an opening, one takes it.
Edit to reflect additional question on charged pions in comment: Same thing! You know something more about the charged pions, that they are symmetric, so you symmetrize: The resulting state of them, then, is orthogonal to $\pi^0 \pi^0$, and still normalized, and by symmetry avoids the antisymmetric isovector, so being isotensor and isoscalar, $\sqrt{1/3} ~\mid  2,0\rangle +\sqrt{2/3} ~\mid  0,0\rangle$, yields $\sqrt{2/3}$ in the amplitudes and 2/3 in the rate, twice as for neutral pions.
Edit/Appendix to the most primitive  shortcut to the answer: Represent the states in |a> with $\Uparrow$ for isospin 1/2 and $\uparrow$ for isospin 1, so the $D^+$ and the $\pi^+$, so $|a\rangle=|\Downarrow 0 0\rangle$. There are 5 states at $M=-1/2$, involving $\Downarrow 00~;~ \Downarrow \uparrow\downarrow~;~ \Downarrow \downarrow\uparrow~;~ \Uparrow\downarrow 0~;~\Uparrow0\downarrow$ .  Construct the isodoublet with $M=-1/2$, annihilated by the lowering operator. 
Discard the two states antisymmetric in the two pions, and the one that, although symmetric, $\mid \Uparrow(0\downarrow+\downarrow 0)\rangle $, still cannot be in the isodoublet, since the lowering operator on it will generate states impossible to cancel by the lowering of the remaining three terms. 
The remaining 3-state combination annihilated by the lowering operator, is, then, uniquely, 
$$
\mid 1/2,-1/2\rangle=\mid \Downarrow(\uparrow\downarrow+\downarrow\uparrow-00)  \rangle/\sqrt{3},
$$
and normalized; hence $\langle a\mid 1/2,-1/2\rangle=-1/\sqrt{3} $ . 
