How many ways can we write digits (1) in place of the letters so that we get a correct subtraction? (Of course, different letters are replaced by different digits, and zero is not written as the leading digit of a number.)

$$
\overline{FTC}-\overline{ETO}=\overline{KK}
$$
When the place values of the digits are written out, (1) takes the following form:

$$
\begin{gathered}
(100 F+10 T+C)-(100 E+10 T+O)=10 K+K, \\
100(F-E)-(O-C)=11 K \leqq 99 .
\end{gathered}
$$

From this, it is clear that $F-E=1, F=E+1$, and $O-C \geqq 1$. On the other hand, as the difference of two digits, $O-C \leqq 9$, so $11 K \geqq 91$, hence $K=9$, from which $O-C=1, O=C+1$.

Accordingly, for the pairs of letters $E, F$ and $C, O$, we need to write two consecutive digits from $0,1, \ldots, 8$, but $E \geqq 1$, and any of the 5 unused digits can be written in the place of $T$.

If now $E=7$ (and $F=8$), then the digits $0,1, \ldots, 6$ form a continuous sequence. The value of $C$ can be chosen in 6 ways. If $E=1,2, \ldots, 6$, the remaining seven digits always form two separate, continuous sequences (a sequence can also be a single digit). Denoting the number of digits in these sequences by $s_{1}$ and $s_{2}$, respectively, we have $s_{1}+s_{2}=7$, and the value of $C$ can be chosen in $s_{1}-1$ and $s_{2}-1$ ways, respectively, since any digit in the sequence except the largest can be used as $C$. Thus, for each of the 6 values of $E$, $C$ can be chosen in $s_{1}+s_{2}-2=7-2=5$ ways. Therefore, the number of pairs of values for $E, C$ is $6+6 \cdot 5=36$, and taking into account the possible values of $T$, the problem has $36 \cdot 5=180$ solutions.