We will select the 4 digits first and arrange them later.
Out of the 4 digits, one of them should be 2 and one of them should be 3.
2, 3, , .
So, we just need to select the other two digits…
The two digits could be (1,1), (2, 2), (3, 3), (1, 2), (1, 3), or (2, 3).
So, the selection of numbers could be…
2, 3, 1, 1
2, 3, 2, 2
2, 3, 3, 3
2, 3, 1, 2
2, 3, 1, 3
2, 3, 2, 3
Each of these selections could be re-arranged in a number of ways.

So total number of possibilities = (12 + 4 + 4 + 12 + 12 + 6) = 50 ways.
Alternate method:
(Arrangements with at least one 2 and one 3) = (All possible arrangements) - (Arrangements with either 1 or 2) - (Arrangements with either 1 or 3) + (Arrangements with only 1)
Think why we need to add (Arrangements with only 1)!

_, _, _, _
(All possible arrangements) = 3⁴
Each blank could be any one of 1, 2 or 3.
(Arrangements with either 1 or 2) = 2⁴
Each blank could be any one of 1 or 2.
(Arrangements with either 1 or 3) = 2⁴
Each blank could be any one of 1 or 3.
(Arrangements with only 1) = 1
Each blank is filled with 1.
(Arrangements with at least one 2 and one 3) = (All possible arrangements) - (Arrangements with either 1 or 2) - (Arrangements with either 1 or 3) + (Arrangements with only 1)
(Arrangements with at least one 2 and one 3) = 3⁴ - 2⁴ - 2⁴ + 1
(Arrangements with at least one 2 and one 3) = 81 - 16 - 16 + 1
(Arrangements with at least one 2 and one 3) = 82 - 32 = 50