With hybridisation, you only hybridise the number of orbitals you need i.e. that are filled. So in methane, with four bonds, you have four electron pairs which need to fit somewhere. So you need four orbitals. Therefore we say that three 2p orbitals hybridise with the 2s orbital, making four sp3 hybrids (the superscripts denoting the number of each pure orbitals involved) each with 1 electron in them. Note that we had to invoke the idea of hybridisation here, since the electron configuration of carbon is 1s22s22p2, and this would only give us two bonds. We know experimentally that this isn't true (in other words, methane exists!) and so have to conjure a trick to work around it.
Alternatively, let us look at the CH3+ cation. This has three bonds, and so three electron pairs. This time we need to invoke the idea of hybridisation to give us three orbitals. So we say that two 2p orbitals hybridise with the 2s orbital, to give three sp2 orbitals. This leaves us with one p-orbital that we didn't hybridise, which would be a non-bonding orbital (in fact, empty), and so this lies perpendicular to the plane of the molecule.
For another example, to clarify further, consider ammonia. Now ammonia has three bonds, but it also has a lone pair. We count the lone pairs in our hybridisation scheme. Therefore we also need to hybridise to obtain four orbitals, like in methane, despite only having three bonds. Thus ammonia is also sp3, mixing three 2p and one 2s orbital. Note however that in contrast to methane, three of these hybrid orbitals are singly occupied, and one is completely filled (the one which contains the lone pair).
Now, we can look at cases by which orbitals do not hybridise.
For heavy atoms such as lead and bismuth (the same groups as carbon and nitrogen respectively) hybridisation does not occur. This is referred to as the inert pair effect. In this case inert pair refers to the unwillingness of the 6s2 orbital to undergo hybridisation. This is because heavy atom self-self bonds i.e. Pb-Pb are incredibly weak. We can imagine that some energy is expended in the hybridisation process because we are mixing s orbitals with p-orbitals of the same principal quantum number, of which the p are always higher in energy (said to be non-degenerate). So the hybrid orbitals are higher in energy than the pure s, and ever so slightly lower than the pure p (but not by much, as the p-character of an sp3 hybrid is 75%). So, the energy gained in forming a heavy atom bond such as Pb-Pb isn't enough to warrant hybridisation, since hybridisation requires energy.
Notice that I've used anthropogenic words to explain this i.e. (we hybridise as opposed to the atoms hybridise). This is deliberate. It's unclear whether hybridisation really occurs or not; indeed more sophisticated theories can be used to explain the phenomena. Hybridisation is more of a caveat used to explain something we couldn't readily explain with the previous level of theory. It's perhaps telling that we only invoke hybridisation when we need to. It's a characteristic of a bad theory when that happens. The idea is invoked readily in chemistry however, so you would do well to familiarize yourself with the terminology.If you're interested more in this topic, look up molecular orbital theory for a more nuanced explanation.