Concatenating the three prices in order gives an arrangement of the seven digits; count arrangements, then the ways to cut one into three prices

Each arrangement is cut into three nonempty blocks of at most $4$ digits: the block lengths are permutations of $1,2,4;$ $2,2,3;$ or $1,3,3$

Concatenating the three guessed prices in order produces an arrangement of the seven given digits, and each guess is recovered uniquely from an arrangement together with a way to cut it into three consecutive nonempty blocks of at most four digits each (prices run from $\$1$ to $\$9999,$ and no price can start with $0$ here since every digit is $1$ or $3$). There are $\frac{7!}{4!\,3!} = 35$ arrangements of four $1$s and three $3$s.

The ordered block lengths are the ways to write $7$ as an ordered sum of three parts between $1$ and $4:$ the permutations of $(1, 2, 4),$ $(2, 2, 3),$ and $(1, 3, 3),$ giving $6 + 3 + 3 = 12$ cuts for each arrangement.

The total is $35 \cdot 12 = 420.$