Let b,r,w be the number of blue bricks,red ones and white ones

\(\left(b,r,w\in Z^+;b+r+w=50;b< r< w;w=11b\right)\).We have :

\(\circledast b+r+w=50\Rightarrow b+w< 50\)

\(\circledast w>r\Rightarrow b+w>r\Rightarrow2\left(b+w\right)>r+b+w\Rightarrow b+w>25\)

\(\Rightarrow25< 12b< 50\Rightarrow2< b\le4\)

If b = 3,then w = 33 and r = 50 - 3 - 33 = 14 (satisfy)

If b = 4,then w = 44 and r = 50 - 4 - 44 = 2 (absurd)

Hence,(b,r,w) = (3 ; 14 ; 33) and w - r = 19

So,the answer is C

The answer is C: 19

There are: 33 bricks of white, 3 bricks of blue (33:11=3), 14 bricks of red (50-33-3=14)

There are: 33-14=19 fewer red bricks than white ones.