A question on the strong interaction and charm number

Question

I know that the particle $P_c(4380)^+$, has quark content $\bar{c}cuud$. Furthermore I know that the reaction $K^-+P_c(4380)^+\rightarrow K^-+J/\psi+p$, is strong and the quark content on the left hand side is $s\bar{u}+c\bar{c}+uud$.

I have two questions:

1) Intuitively one would expect that as this is a strong interaction that all the quarks on the right hand side should match up with all the quarks on the left hand side , correct ? Which makes it obvious (say we didn't know and we were trying to figure it out ) that the quark content of the $P_c(4380)^+$ particle is $\bar{c}cuud$.

2) The definition of the charm number is $C=n_c-n_{\bar{c}}$, so on the right hand side this gives zero . How can we use this number for determining how many charm quarks are involved considering , you could have zero charm, anticharm quarks or any number of them ( as long as it's the same amount for both) and it will still give C=0.

Answer 1

The part you're missing is that the $P_c(4380)$ was observed in $\Lambda_b$ decays: $$ \Lambda_b \to P_c(4380) K^- \to J/\psi p K^- $$

The presence of a kaon and a $J/\psi$ in the final state tells you this decay involves a $b \to c\overline{c}s$ transition. The $\Lambda_b$ is the lightest $b$-baryon and there is no quark with a mass between that of the $c$ and $b$ quarks, so any intermediate state decaying to the $J/\psi$ must contain $c\overline{c}$. Therefore any $X \to J/\psi p$ observed in this system must be a $c\overline{c}uud$ state and any $X \to J/\psi K^-$ must be $c\overline{c}s\overline{u}$ (the $B^+$ is too heavy).

So to answer your questions: 1) yes, but it's the other way around: we know this is allowed as a strong decay because the quark content doesn't change and 2) yes, all that $C=0$ tells you is that there are an equal number of $c$ and $\overline{c}$ quarks.

^{From arXiv:1507.03414.}