6 boxes with 4 blue balls and 5 red balls How many ways can we arrange the balls, obviously irrespective of order? I could figure out a basic example where we have $6$ boxes and one group of $4$ and one group of $2$, with an answer of $\displaystyle\frac{6!}{4!2!}$, but I'm having trouble with how to approach this. It's not a problem in my textbook but the author says that solving something like this is much more complex in detail but the ideas are not more complex. But he does not discuss it any further saying it seldom arises in practice. 
 A: Hint: Use Stars and Bars (please see Wikipedia) to find the number of ways to arrange the reds. Use Stars and Bars to find the number of ways to arrange the blues. Multiply.
The idea works for any number of colours, where the number of balls of each colour, and the number of boxes, are specified.
A: Let $x_k$ denote the number of blue balls placed in box $k$.  Then 
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 4$$
which is an equation in the nonnegative integers.  The number of solutions of the equation is the number of ways we can place five addition signs in a row of four ones, which we can do in 
$$\binom{4 + 5}{5} = \binom{9}{5}$$
since we must select which five of the nine symbols (four ones and five addition signs) will be addition signs.
Let $y_k$ denote the number of red balls placed in box $k$.  Then 
$$y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 5$$
which is an equation in the nonnegative integers.  The number of solutions is the number of ways five addition signs can be placed in a row of five ones, which is 
$$\binom{5 + 5}{5} = \binom{10}{5}$$
Hence, the number of ways four indistinguishable blue balls and five indistinguishable red balls can be placed in six boxes is 
$$\binom{9}{5}\binom{10}{5}$$
