jmc

Ignoring the different colors of paper, we can put the stickers on the sheets of paper in the following groups: $$\begin{array}{rl}& (8,0,0,0)\\ & (7,1,0,0)\\ & (6,2,0,0)\\ & (6,1,1,0)\\ & (5,3,0,0)\\ & (5,2,1,0)\\ & (5,1,1,1)\\ & (4,4,0,0)\\ & (4,3,1,0)\\ & (4,2,2,0)\\ & (4,2,1,1)\\ & (3,3,2,0)\\ & (3,3,1,1)\\ & (3,2,2,1)\\ & (2,2,2,2).\end{array}$$For each of these combinations, we will list how many distinct ways there are to put the groups of stickers on the different sheets of paper.

$\bullet$ For $(8,0,0,0),$ there are $\frac{4!}{3!}=4$ ways.

$\bullet$ For $(7,1,0,0),$ we have $\frac{4!}{2!}=12$ ways.

$\bullet$ For $(6,2,0,0),$ there are $\frac{4!}{2!}=12$ ways.

$\bullet$ For $(6,1,1,0),$ there are $\frac{4!}{2!}=12$ ways.

$\bullet$ For $(5,3,0,0),$ we have $\frac{4!}{2!}=12$ ways.

$\bullet$ For $(5,2,1,0),$ there are $4!=24$ ways.

$\bullet$ For $(5,1,1,1),$ there are $\frac{4!}{3!}=4$ ways.

$\bullet$ For $(4,4,0,0),$ we have $\frac{4!}{2!2!}=6$ ways.

$\bullet$ For $(4,3,1,0),$ there are $4!=24$ ways.

$\bullet$ For $(4,2,2,0),$ there are $\frac{4!}{2!}=12$ ways.

$\bullet$ For $(4,2,1,1),$ we have $\frac{4!}{2!}=12$ ways.

$\bullet$ For $(3,3,2,0),$ there are $\frac{4!}{2!}=12$ ways.

$\bullet$ For $(3,3,1,1),$ there are $\frac{4!}{2!2!}=6$ ways.

$\bullet$ For $(3,2,2,1),$ we have $\frac{4!}{2!}=12$ ways.

$\bullet$ For $(2,2,2,2),$ there are $\frac{4!}{4!}=1$ way.

In total, there are $$4+12+12+12+12+24+4+6+24+12+12+12+6+12+1=\boxed{165}$$ways for Henry's brother to put the stickers on the sheets of paper.

Notice that this answer equals $\left(\genfrac{}{}{0ex}{}{11}{3}\right).$ Is this a coincidence?