Answer
All C-C bonds are formed using $sp^{3}$ hyrbid orbitals while C-H bonds are formed with an $sp^{3}$ hybrid orbital, from the carbon atom, and an s orbital, from the hydrogen atom.
C-C bond: $sp^{3}-sp^{3}$
C-H bond: $sp^{3}-s$
There are 10 sigma bonds present in propane.
Work Step by Step
The first step is to identify the hybridization of every atom in the molecule. To do this, all you have to do is count the number of bonds it has. Also, keep in mind that double and triple bonds are counted as 1 and that lone pairs count as 1 "bond."
4 bonds = $sp^{3}$ hybridized
3 bonds = $sp^{2}$ hybridized
2 bonds = $sp$ hybridized
For example, the carbon atom found in $CH_{4}$ has four bonds; therefore, the carbon atom is $sp^{3}$ hybridized. On the other hand, hybridization is not need for hydrogen since it can only form one bond. All hydrogen atoms will always have a 1s orbital. With this, we can figure out the orbitals used to form the C-H bonds: $sp^{3}-s$.
To find the number of sigma bonds all you have to do is count the number of bonds the molecule has. Single, double and triple bonds all have one sigma bond.
