Spin of 3 particles I am trying to decompose the isospins of a three particle state using Clebsch-Gordan coefficients such as: 
$|1,1\rangle \otimes |1/2,-1/2\rangle \otimes |1,0\rangle$ 
Decomposing the first two states gives: 
$|1,1\rangle \otimes |1/2,-1/2\rangle = \sqrt{\frac{1}{3}}|3/2,1/2\rangle + \sqrt{\frac{2}{3}}|1/2,1/2\rangle$
And then these combined with the third state give:
$|3/2,1/2\rangle \otimes |1,0 \rangle = \sqrt{\frac{3}{5}}|5/2,1/2\rangle + \sqrt{\frac{1}{15}}|3/2,1/2\rangle - \sqrt{\frac{1}{3}}|1/2,1/2\rangle$ 
$|1/2,1/2\rangle \otimes |1,0 \rangle = \sqrt{\frac{2}{3}}|3/2,1/2\rangle + \sqrt{\frac{1}{3}}|1/2,1/2\rangle$
When I combine these all together I get:
$|1,1\rangle \otimes |1/2,-1/2\rangle \otimes |1,0\rangle = \sqrt{\frac{1}{5}} |5/2,1/2\rangle + \frac{10+\sqrt{5}}{15} |3/2,1/2\rangle + \frac{-1+\sqrt{2}}{3}|1/2,1/2\rangle$
Which has to be incorrect as this state is not normalised. Basically my question is, what am I doing wrong? 
Edit: What I'm attempting to calculate is amplitudes for processes like $\Lambda p \to \Lambda p \pi^0$ using isospin states for all of the particles. 
 A: Using software, I get:

for ket in Ket(1,1)*Ket(0.5,-0.5)*Ket(1,0):
      ...:     print unicode(ket)
      ...:
      ...: 
√（2∕9) |½, ½⟩
-⅓|½, ½⟩
⅔|1½, ½⟩
1∕25 |1½, ½⟩
√（⅕) |2½, ½⟩

whose coefficients add (in quadrature) to unity. So the question is, why can't you add the two $|\frac 1 2, \frac 1 2\rangle$ kets coherently (and likewise for the two $|\frac 3 2, \frac 1 2\rangle$ kets)?
See the comments--there are different mixed symmetry representations in the final answer with the same multiplicity.
A: So the key point here is to realize that the coupling 
$$
1\otimes \frac{1}{2}\otimes 1
$$
will contain some final $J$ values more than once.  Indeed 
$$
1\otimes \frac{1}{2}=\frac{3}{2}\oplus \frac{1}{2} \tag{1}
$$ 
and coupling this to $1$ will 
produce, for instance,  two types of states with final $J=\frac{1}{2}$, depending on the intermediate $J_{12}$ value.  Thus, copy will come from the $J_{12}=\frac{3}{2}$ and the other will come from the $J_{12}=1$ states of (1).
To be systematic write
$$
\vert 1,1\rangle
\vert\textstyle\frac{1}{2},-\frac{1}{2}\rangle
=\frac{1}{\sqrt{3}}\vert \frac{3}{2}\frac{1}{2}\rangle + \sqrt{\frac{2}{3}}
\vert\frac{1}{2}\frac{1}{2}\rangle\, .\tag{2}
$$
Coupling the $\frac{3}{2}$ state to $1$, the part proportional to final $J=\frac{1}{2}$ using the CG $C_{3/2,1/2;1,0}^{1/2,1/2}=-\frac{1}{\sqrt{3}}$ yields
$\vert \frac{1}{2}\frac{1}{2};J_{12}=\frac{3}{2}\rangle$ with
$$
\langle \textstyle\frac{1}{2}\frac{1}{2};J_{12}=\frac{3}{2}\vert
1,1;\frac{1}{2},\frac{1}{2} ; 1,0\rangle
=-\frac{1}{3}
$$
but going through the $J_{12}=\frac{1}{2}$ produces a different $J=\frac{1}{2}$ state with
$$
\langle \textstyle\frac{1}{2}\frac{1}{2};J_{12}=\frac{1}{2}\vert 
1,1; \frac{1}{2},\frac{1}{2} ; 1,0\rangle
=+\frac{\sqrt{2}}{3}
$$
You can actually check that $\vert\textstyle\frac{1}{2}\frac{1}{2};J_{12}=\frac{1}{2}\rangle$ is really different from 
$\vert\textstyle\frac{1}{2}\frac{1}{2};J_{12}=\frac{3}{2}\rangle$ by computing
their explicitly expressions in terms of $j_1=1,j_2=\frac{1}{2}, j_3=1$ states; you will see that these are distinct linear combinations of the $j_1=1,j_2=\frac{1}{2}, j_3=1$ states.
