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Solution

(interpreting “contains 1” as **exactly one** ‘1’, repetitions allowed):** Case-1: ‘1’ in the thousand’s place → remaining $3$ places from $\{0,\dots,7\}\setminus\{1\}$ with repetition: $7^3=343$ ways. Case-2: ‘1’ in any one of the last three places ($3$ choices). Thousand’s place from $\{2,\dots,7\}$ ($6$ ways). Remaining two places from $\{0,\dots,7\}\setminus\{1\}$ with repetition: $7^2=49$ ways. Total $=343+3\cdot6\cdot49=343+882