A string of length 7 is formed using the letters A, B, C such that it begins with A, no two adjacent letters are the same, and all three letters…

Strings of length $7$ starting with $A$ with no two adjacent letters equal: each of the remaining $6$ positions has exactly $2$ choices (any letter except the previous one), giving $2^6=64$ strings. Now remove those in which a letter is missing. The letter $A$ is present at the start, so the missing letter is $B$ or $C$. If $C$ is missing, the string must alternate $ABABABA$, which is exactly $1$ string. Likewise if $B$ is missing, only $ACACACA$ works. These two cases cannot overlap, because a string using only $A$ cannot satisfy the adjacency rule. Therefore the required count is $64-1-1=62$. A computer enumeration of all $3^7$ strings confirms $62$.