The only factors of 33 are 1,3,11,33

N × U must be one of the factors and M + B + E + R must be another.

N × U can only be 3, 

for if it were 1, M + B + E + R would have to be 33 -- impossible!
and if it were 11, M + B + E + R would have to be 3 -- also impossible! 
and if it were 33, M + B + E + R would have to be 1 -- also impossible!

So N and U are 1 and 3, and M + B + E + R = 11

So we must pick four digits from 0,2,4,5,6,7,8,9
that have sum 11.

The only ones that have sum 11 are the smallest four, 
0 + 2 + 4 + 5 = 11

So the first way is

1 × 3 × (0 + 2 + 4 + 5) = 33

The 1 and 3 can be arranged in 2! = 2 ways
The 0,2,4, and 5 can be arranged in 4! = 24 ways


Answer:   2! × 4! = 2 × 24 = 48

Edwin