In order to be divisible by 55, it must be divisible by both 5 and 11. That means the last digit must be 0 or 5 and the sum of the 1st, 3rd and 5th digits less the 2nd and 4th digits must be divisible by 11.
Let me count the ways.
If the digits are 01234 then they sum to 10. 0 is the last digit and the odd places (2 and 4) must sum to 5.
this gives us two choices for the odd places times 2 ways to place them times 2 ways to place the 1st/3rd digit.
Altogether, 8 possibilities.
They are 12430, 13420, 42130, 43120, 21340, 24310, 31240, 34210
If the digits are 12345 then they sum to 15. 0 possibilities.
If the digits are 23456 then they sum to 20. 5 is the last digit and the odd places (2 and 4) must sum to 10. This can only be done in 1 way(6+4) times 2 ways to place them times 2 ways to place the 1st/3rd digit.
Altogether, 4 possibilities.
If the digits are 34567 then they sum to 25. 5 is the last digit and the odd places (2 and 4) must sum to (25-11)/2 = 7. This can only be done in 1 way(3+4) times 2 ways to place them times 2 ways to place the 1st/3rd digit.
Altogether, 4 possibilities.
If the digits are 45678 then they sum to 30. 5 is the last digit and the odd places (2 and 4) must sum to 30/2 = 15. This can only be done in 1 way(7+8) times 2 ways to place them times 2 ways to place the 1st/3rd digit.
Altogether, 4 possibilities.
If the digits are 56789 then they sum to 35. 5 is the last digit and the odd places (2 and 4) must sum to (35-11)/2 = 12. 0 possibilities.
Final answer = 8 + 4 + 4 + 4 = 20 in the class of 55
Let me count the ways (spoiler)
