first place should be occupied by either $3$ or $4.$
Case 1 : First place is occupied by the digit $4$
$4$ _ _ _
now in the set from where we can pick numbers is left with $=\{2,2,3,3,3,4,4,4\} $
if we got $3$ of each digit(which are $2, 3$ and $4$) then number of ways by each of those blanks can be
filled in are $3$ coz we have $3$ choices of digits: $\text{pick } 2, 3$ or $4.$
But we do not have just enough $2's$ to fill all those $3$ spaces with the digit $2.$
$\therefore$ we need to subtract this case where number would be $4222$.
So, total numbers obtained using the numbers in our current set $=1 \times 3 \times 3 \times 3 - 1 = 26$.
The first one is for the digit $4,$ coz its fixed for this case; the subtracted one is for the case $4222$ that can't be made possible.
Case 2: First place is occupied by the digit $3$
$3$ _ _ _
now in the set from where we can pick numbers is left with $=\{2,2,3,3,4,4,4,4\}$
we have enough $4's$ here but lack $3's$ and $2's$ $\therefore$, the cases we need to subtract are $3222$ and $3333$
So, total numbers obtained using the numbers in our current set $=1 \times 3 \times 3 \times 3 - 2 = 25$
both cases are independently capable of giving us the answer, we have =$ 26 + 25 = 51.$
Hence answer is option B.