In how many ways can we place $3$ red and $4$ blue balls into $3$ indistinguishable boxes so that no box is empty? In how many ways can we place $3$ red and $4$ blue balls into $3$ indistinguishable boxes so that no box is empty?
What if the boxes can be empty?

Using the Polya Enumeration Theorem we get for the two cases using the
cycle index of the symmetric group with empty boxes
$$Q_1 = [R^3 B^4] Z\left(S_3;
(1+R+R^2+R^3)\times(1+B+B^2+B^3+B^4)\right)$$
and without
$$Q_2 = [R^3 B^4] Z\left(S_3; -1 +
(1+R+R^2+R^3)\times(1+B+B^2+B^3+B^4)\right).$$
Now the cycle index is
$$Z(S_3) = 1/6\,{a_{{1}}}^{3}+1/2\,a_{{2}}a_{{1}}+1/3\,a_{{3}}.$$
Doing the substitution we find
$$\bbox[5px,border:2px solid #00A000]{
Q_1 = 28 \quad\text{and}\quad Q_2 = 18.}$$
If  we want  to do  these by  hand,  here is  an example.  We use  the
alternate form
$$[R^3 B^4] Z\left(S_3;
\frac{1}{1-R}\frac{1}{1-B}\right).$$
We get from the first term of the cycle index
$$[R^3 B^4] \frac{1}{6}
\frac{1}{(1-R)^3}\frac{1}{(1-B)^3}
= \frac{1}{6} {3+2\choose 2} {4+2\choose 2} = 25.$$
We get from the second term
$$[R^3 B^4] \frac{1}{2}
\frac{1}{1-R^2}\frac{1}{1-B^2}
\frac{1}{1-R}\frac{1}{1-B}
= \frac{1}{2} (1+1)\times (1+1+1) = 3.$$
Here  we have  e.g. for  the  coefficient on  $B^4$ the  possibilities
$(B^2)^2 (B^1)^0,$ $(B^2)^1 (B^1)^2$ and $(B^2)^0 (B^1)^4.$
At last we get from the third term
$$[R^3 B^4] \frac{1}{3}
\frac{1}{1-R^3}\frac{1}{1-B^3} = 0.$$
Add these to obtain $25+3=28.$
