Classification of surfaces

The Classification Theorem for surfaces says that a compact connected surface $$M$$ is homeomorphic to $$S^2\# (\#_{g}T^2)\# (\#_{b} D^2)\# (\#_{c} \mathbb{R}P^2),$$ so $$g$$ is the genus of the surface, $$b$$ the number of boundary components and $$c$$ the number of projective planes.

From there, it is easy to compute $$\chi(M)=2-2g-b-c$$.

Nevertheless, I have read another statement of The Classification Theorem that states that a compact connected surface is determined by its orientability (yes/no), the number of boundary components and its Euler characteristic.

I do not understand how is it possible to know the decomposition of $$M$$ as a connected sum by knowing that. By knowing $$b$$, there are still two variables, $$c$$ and $$g$$ which have to be known from $$\chi(M)$$, and orientability only tells us if $$c=0$$ or $$c\geq 1$$. Can someone help me, please?

Since $$\mathbb{R}P^2\# \mathbb{R}P^2\#\mathbb{R}P^2\cong \mathbb{R}P^2 \# T^2$$, $$c$$ and $$g$$ are not uniquely determined: if $$c\geq 3$$, you can subtract $$2$$ from $$c$$ and add $$1$$ to $$g$$ and get the same surface, or if $$c,g\geq 1$$, you can subtract $$1$$ from $$g$$ and add $$2$$ to $$c$$.
Note, though, that the first operation can always be used to get a connected sum presentation where $$c\leq 2$$. If you impose the additional restriction that $$c\leq 2$$, then $$c$$ and $$g$$ can be uniquely determined and can be calculated from the data you mention. If the surface is orientable, then $$c=0$$ and then you can just solve for $$g$$. If the surface is not orientable, then you can determine whether $$c=1$$ or $$c=2$$ since $$c$$ must have the same parity as $$\chi(M)+b$$. Once $$c$$ is determined, you can solve for $$g$$.
If I read correctly you are trying to determine $$c$$ and $$g$$, given a compact connected surface $$M$$ for which you know $$b$$ the number of boundary components, $$\chi(M)$$ the Euler characteristic the and whether or not $$M$$ is orientable.