# Spin-parity states of $^{17}O$ and $^{19}O$

I am trying to understand a problem in a subnuclear physics course. The question is the following:

The biggest difference between the low-lying states of $$^{17}O$$ and $$^{19}O$$ is that $$^{19}O$$ has two extra states with spin-parity $$\frac{3}{2}^+$$ and $$\frac{9}{2}^+$$. Show that these are the result of the configuration $$(d_{5/2})^3$$ and are therefore not expected in $$^{17}O$$.

What I have done:

I have used the shell model to confirm that the states with an unpaired neutron are $$(d_{5/2})^1$$ for $$^{17}O$$ and $$(d_{5/2})^3$$ for $$^{19}O$$, while the protons are always paired.

In the event of unpaired nucleons, the total spin can be all values satisfying $$|j_p - j_n| < I < j_p + j_n$$. But in this case of paired protons, how can there even be multiple states? And how can it be different between two configurations which both have $$j = 5/2$$?

I have a solution to the problem, that I don't understand. It goes like this, define $$j_1 = j_2 =j_3 = 5/2.$$ Now, write down all integer values between $$j_1 + j_2$$ and $$|j_1 - j_2|$$. Set $$j' = j_1 + j_2$$. Now write down all the values between $$j' + j_3$$ and $$|j' - j_3|$$. We can see that in this second list we find the state that only exist for $$^{19}O$$.

I don't understand the solution. Any help would be much appreciated!