\(\displaystyle (((1/6)*s)^{10})*((1-s^6)/(1-s))^{10}\)

Which is equal to

\(\displaystyle ((1/6)^{10})*(1-10s^6+...)*(1+10s+...)\)

The last line should be multiplied by \(\displaystyle s^{10}\).

By the simplest form of the binomial theorem we have

\(\displaystyle (1-s^6)^{10}=1-\binom{10}{1}s^6+\binom{10}{2}s^{12}+\dots\) (1)

Also, by the last formula in

this Wiki section, we have

(2)

We have \(\displaystyle s^{10}\) that comes from the first factor \(\displaystyle ((1/6)*s)^{10}\), so the rest of the expression must contribute \(\displaystyle s^{17}\) towards \(\displaystyle s^{27}\). We can get \(\displaystyle s^{12}\) from (1) and \(\displaystyle s^{5}\) from (2); \(\displaystyle s^{6}\) from (1) and \(\displaystyle s^{11}\) from (2); and \(\displaystyle s^{0}\) from (1) and \(\displaystyle s^{17}\) from (2). The product of the corresponding coefficients make the three terms in the expression

\(\displaystyle \binom{10}{2}\binom{14}{5}-\binom{10}{1}\binom{20}{11}+\binom{26}{17}\)