Puzzle on Ruby - Part 2

Let's reviews back the previous last puzzle on "Puzzle on Ruby - Part 1":

Puzzle:

ABC + BDE = CEF

Condition #1: Every alphabet is represent a digit between 0 to 9.
Condition #2: Whenever a digit has been used, that digit cannot be used anymore for other alphabets.
Condition #3: 8 * ( ABC - BDE ) = CEF

Answer:
A = 5, B = 4, C = 9, D = 2, E = 7, F = 6 because

549 + 427 = 976 and
8 * (549 - 427) = 976

Hopefully, you also got the answer correctly.

~..~..~ The End ~..~..~

Okay! Let's improve puzzle on Part 1 to become as below:

Puzzle:

GHIJ + HABC + IBDE = JCEF

Condition #1: Every alphabet is represent a digit between 0 to 9.

Total is 175725 solutions available when only condition #1 is applied. (Pretty easy, right? You could find a lot of different solutions easily.)

Condition #2: Whenever a digit has been used, that digit cannot be used anymore for other alphabets.

Total is 80 solutions available when condition #1 and #2 are applied. (Oh Gosh! This is pretty hard. You may not even find one solution out through many many trials and errors)

How about giving one more condition to the puzzle? Will it be more harder to be solved?

Condition #3: GHIJ, HABC, IBDE, and JCEF are even numbers.

Total is 1 solution available. (And...it's pretty easy to solve.)

As you can see, the number of solutions is limited by the content of puzzle's conditions. And the more conditions applied to the puzzle, the less solutions the puzzle can have.

The level of a puzzle depends on the conditions as well. Level of a puzzle is proportional to the number of conditions applied to that puzzle. However, this statement is not always true. Like the puzzle above, when the third condition applied, the puzzle becomes simple and lost its challenging characteristic anymore.

Anyway, you may be curious and want to find the only one solution out. Good luck to you!