Very interesting question Mariah!
Please see the image (solution structure) below to review along with the explanation.
The solution can be structured in 7 'selection conditions' (SC 1 to SC 7)
SC1: All five Ls (letters) are the same
Only possible for "A" = 1 letter => Number of ways = 1C1 = 1 way .............................(1)
SC2: Only 4 Ls are the same (Only possible with A and B) and one is different (1 out of 4 remaining, after A or B is selected)
=> Number of ways = 1*4C1 + 1*4C1 = 8 (Same as 2C1*4C1 = 8).............(2)
SC3: Only 3 Ls are the same (Possible with A, B and C) and two are diff.( 2 out of 4 remaining, after A or B or C is selected)
=> Number of ways = 1*4C2+1*4C2+1*4C2 = 18 (same as 3C1*4C2 = 18) .......(3)
SC4: 3 Ls are the same (Possible with A, B and C) and two others are same (Possible with A, B, C and D => 2 letters of 1 letter type from 3 remaining letter types)
=> Number of ways = 1*3C1+1*3C1+1*3C1 = 9 (Same as 3C1*3C1 = 9) .........(4)
SC5: 2 Ls are the same (Possible with A, B, C and D) and three are diff (same logic as above)
=> Number of ways = 1*4C3+1*4C3+1*4C3+1*4C3 = 16 (same as 4C1*4C3 = 16) .......(5)
SC6: 2 Ls are the same (Possible with A, B, C and D) and 2 other Ls are the same (Possible with A, B, C and D)and one different letter (same logic as above)
=> Number of ways = 1*3C1*3C1 +1*2C1*3C1 +1*1C1*3C1 = 18 (Same as 4C2*3C1= 18) .....(6)
[SC6 is a little tricky because we need to avoid repetition as shown above]
SC7: All 5 Ls are diff (Possible only with A, B, C, D and E)
=> Number of ways = 5C5 = 1 ...................................................(7)
Adding the number of ways from the 7 SCs = 1+8+18+9+16+18+1 = 71
Hope this helps...
