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**Shanks** (1)each digit has two options:0,1. therefore the number of codes can be stored is $\displaystyle 2^{10}$.

(2)Classification:

case (a): only one S in the 6 letters

then the 6 letters are one S, two T, three E. By the multiplication principle, the numbers of arrangements:$\displaystyle {6\choose 1}{5\choose 2}$.

case (b): two S in the 6 letters

then the 6 letters are possibly (bi): two S, one T, three E; (bii): two S, two T, Two E.

for case(bi): the number of arrangements:$\displaystyle {6\choose 1}{5\choose 2}$.

for case(bii): the number of arrangements: $\displaystyle {6\choose 2}{4\choose 2}$.

Sum these numbers up to get the total number of arrangements.