Find the number of 7-digit strings formed using only the digits 1,2,3, in which each of the three digits appears at least once and the sum of all the…

Let $(a,b,c)$ be the numbers of $1$s, $2$s, $3$s, with $a+b+c=7$ and each at least $1$. The digit sum is $a+2b+3c\equiv a+2b\equiv a-b\pmod 3$, so we need $a\equiv b\pmod3$. The admissible triples are $(1,1,5),(1,4,2),(4,1,2),(2,2,3),(3,3,1)$, contributing $\frac{7!}{a!\,b!\,c!}=42,105,105,210,140$ arrangements respectively, which total $602$. As a cross-check, a complete brute-force enumeration of all $3^{7}=2187$ strings, keeping those that use all three digits with digit sum divisible by $3$, returns exactly $602$ strings. The answer is $602$. The congruence argument is shorter than a roots-of-unity filter, and the surjectivity condition is built in because every composition has each part at least one. A common slip is to forget that the multinomial coefficient must be used for arrangements, not just the number of compositions, which would give only five.