I.  Those containing 0 and 5

The remaining 3 are arrangements of
3 from the set {2,2,3,3,3}

These are 7 of those:

223,232,233,322,323,332,333

Consider any one of those as  
    XYZ
 
A.  The 5 comes before the 0

   1. The 5 precedes the X. (5XYZ)
      There are 4 positions for the 0.
   2. The 5 is between the X and Y.
      (X5YZ)
      There are 3 positions for the 0. 
   3. The 5 is between the Y and Z.
      (XY5Z)
      There are 2 positions for the 0.
   4. The 5 is after the Z.
      (XYZ5)       
      There is 1 position for the 0.
      
      A total of 4+3+2+1 = 10 ways       

B.  The 0 comes before the 5.

   1. The 0 is between the X and Y.
      (X0YZ)
      There are 3 positions for the 5.
   2. The 0 is between the Y and Z.
      (XY0Z)      
      There are 2 positions for the 5.
   3. The 0 is after the Z.
      (XYZ0)
      There is one position for the 5.

      A total of 3+2+1 = 6 ways 

So there are 10+6 = 16 ways to insert 
a 0 and a 5 in each of those 7 sequences 
of three.

So there are 16×7 = 112 ways to have both 
a 0 and a 5.

II.  Those containing a 5 or a 0 but not both

The remaining 4 are arrangements of
3 from the set {2,2,3,3,3} 

which are these 10 ways:

2233,2323,2332,3223,3232,3322,
2333,3233,3323,3332 

There are 5 positions in each to insert a 5.
There are 4 positions in each to insert a 0

That's 9 ways to insert a 0 or 5 in each of 
those 10 arrangments.  That accounts for 9×10 
or 90 ways.1

III.  Those containing only 2's and 3's

There are 10 of those

22333, 23233, 23323, 23332, 32233,
32323, 32332, 33223, 33232, 33322.
  
Grand total = 112 from I, 90 from II, and 
10 ways from III = 112+90+10 = 212.

Answer = 212.

Edwin