6. Now we have 1 one-tenth yuan note, 1 one-tenth yuan note, 1 six-tenth yuan note, 5 one yuan notes, and 2 two yuan notes. The number of different amounts (excluding the case of not paying) that can be made is $\qquad$
6. Using: Jiao Shi Ji to match 0 Jiao, 1 Jiao, 2 Zhong different denominations. That is, $(1+i)$ kinds.

Using 5 Jiao can match 0 Jiao, 5 Jiao, which is $(1+1)$ kinds; Yuan. That is, $(5+1)$ kinds;
Using 5 Yuan can match 0 Yuan, 5 Yuan, 10 Yuan, which is $(2+1)$ kinds.
Considering the amounts to be paid are 5.5, 1.5, 2.5, 3.5, 5, 6.5, 7.5, 8.5, 10, 10.1, 10.2, 10.3, 10.5, 10.6, 10.7, 10.8 Yuan, each has 2 methods, so the total number of ways to pay is
$$
\begin{array}{l}
(1+1)(1+1)(1+1)+(5+1)(2+1)-16-1 \\
=127 \text { ( )} .
\end{array}
$$