First a matter of vocabulary: These problems are called

**multi-selections**; and we say the items are

**indistinguishable** not similar.

The number of ways to place $K$ indistinguishable items into $N$ distinct cells is $\dbinom{N+K-1}{N}=\dbinom{N+K-1}{K-1}$

So the four remaining balls can be placed in $\dbinom{4+4-1}{4}$ ways.

To see the logic, suppose we are to place six balls into three boxes, $A,~B,~C$ We could model it as:

$\underbrace \bullet _{~A~}|\underbrace { \bullet \bullet \bullet }_B|\underbrace { \bullet \bullet }_C$ that ix one in A, three in B and two in C.

$\underbrace {\color{white}{\_\_ \_}} _{~A~}|\underbrace { \bullet \bullet }_B|\underbrace { \bullet \bullet \bullet \bullet }_C$ that is none in A, two in B and four in C.

$\underbrace {\color{white}{\_\_ \_}} _{~A~}| \underbrace {\color{white}{\_\_ \_}}_{~B~}|\underbrace { \bullet \bullet \bullet \bullet \bullet \bullet }_C$ ALL six in C.

We see the we can with two 'bars' and six 'bullets' model any containment of six balls into three boxes.

The string $\bullet\bullet\bullet\bullet\bullet \bullet~ |~|$ can be rearanged in $\dfrac{8!}{6!\cdot 2!}=\dbinom{6+3-1}{6}$ ways.

P.S. You can also

read this.