Distinguishing pentanol isomers in 13C DEPT NMR spectroscopy
In an ideal non-moving world at $0\,\mathrm{K}$, every atom would give exactly one peak in NMR spectra. So the actual question is why some of these spectra have less signals. Partly this is due to local symmetry. If, e.g., two methyl groups are attached to the same carbon atom, these two are said to be chemically equivalent, i.e. one can rotate bonds to overlap one over the other. A different example of chemical equivalence is pentan-3-ol which you can rotate around the $\ce{C-OH}$ axis to give an almost identical molecule, putting the $\ce{CH2}$ groups on top of each other and the methyl groups, too.

With that, you did the circling right, and we should expect:


3 spectra with five signals;
3 spectra with four signals; and
2 spectra with three signals.


However we see one spectrum with two signals and are lacking one with four signals. (And technically, if you look at spectrum 2 closely, that could be interpreted as having five signals, not four.)

In a standard $\ce{^{13}C}$-DEPT spectrum (i.e. the ones you showed), carbons from $\ce{CH}$ and $\ce{CH3}$ groups point upward and those from $\ce{CH2}$ groups downward — and there is your answer. There are carbons in the structures that do not have hydrogens bound to them and which thus do not show up in the DEPT spectra.

In case you’re interested, I will give you the appropriate IUPAC names of the pentanol isomers that were measured in the following spoiler tag.



   2-methylbutan-2-ol. Carbon 2 cannot be seen.


   3-methylbutan-2-ol. The two methyl groups at the end are not completely identical and thus give two signals.


   pentan-3-ol  


   pentan-1-ol  


   2,2-dimethylpropan-1-ol. Carbon 2 cannot be seen.


   2-methylbutan-1-ol  


   pentan-2-ol  


   3-methylbutan-1-ol