Condition over Combinations

Question

Given a set of colored balls (multiple balls having the same color is allowed), is there any direct combinatorial approach to determine the total number of ways to select n distinctly-colored balls? Note: n will be less than or equal to the total number of distinct colors.

Example:- Let's say we have 11 colored balls. 2 are Red, 2 are Green, 3 are Yellow, and 4 are Blue. Here, there are 4 distinct colors. So n will be less then or equal to 4.

Consider n = 3. The possible combinations of selecting 3 distinctly-colored balls are {Red, Green, Yellow}, {Red, Green, Blue}, {Red, Yellow, Blue}, {Green, Yellow, Blue}.

Ways to select {Red, Green, Yellow} combination is 2*2*3 = 12. Ways to select {Red, Green, Blue} combination is 2*2*4 = 16. Ways to select {Red, Yellow, Blue} combination is 2*3*4 = 24. Ways to select {Green, Yellow, Blue} combination is 2*3*4 = 24.

Hence the total number of ways to select 3 distinctly-colored balls = 12+16+24+24 = 76.

Another example is when n = 2, the answer will be 4 Red-Green combinations, 6 Red-Yellow combinations, 8 Red-Blue combinations, 6 Green-Yellow combinations, 8 Green-Blue combinations, 12 Yellow-Blue combinations. Total = 4+6+8+6+8+12 = 44.

Last example when n = 4, the answer is 2*2*3*4 = 48.

Answer 1

I doubt there is a nice combinatorial solution.

Suppose the $i$th colour has $b_i$ distinguishable balls and there are $t$ different colours.

Essentially I see three possible approaches: one is the approach you took of finding the possible patterns of colours, multiplying the numbers, and then summing over the patterns.

The second is to consider the coefficient of $x^n$ in the expansion of $$(1+b_1x)(1+b_2x)\cdots(1+b_ix)\cdots.$$

The third is with the recurrence $f(i,c)=f(i,c-1)+b_if(i-1,c-1)$ when $i \le c$ and $f(i,c)=0$ when $i \gt c$, starting with $f(0,c)=1$. You want $f(n,t)$.