# Number of ordered ways to pick 3 balls out of an urn with 2 black and 3 white balls without replacement

The answer is 7: WWW, WWB, WBW, BWW, WBB, BWB, BBW. Is there an elegant way of calculating it in a way that scales to picking M balls out of K black and N white balls?

There isn't really a nice expression for it, no. What you want in general is a sum of binomial coefficients $$\sum_{i = \max(0, M-N)}^{\min(K, M)}\binom{M}{i}$$ where each term is the number of orders in which you can pick $i$ black balls and $M-i$ white balls, and the boundaries of the sum are the given by how many black balls you could possibly draw.

A sum of consecutive binomial coefficients like this has no nice closed form in general. For the three cases $\sum_{i = 0}^M$, $\sum_{i = 0}^{M/2}$ and $\sum_{i = M/2}^M$ (in other words, either there are more than enough balls of both colors, or exactly one of the colors is limited, to exactly $M/2$) there is a nice expression (it's just a power of $2$), and cases close to those can be calculated from there, but that's the only special cases that I know of.

• There ir a common name to the combinatory problems with "limited replacement" depending of the objects? Jul 18 '18 at 9:51

$$\sum_{k=\max\{0,M-N\}}^{k=\min\{K,N\}}\dfrac{ M!}{k!(M-k)!}$$

For each $k$ balls of the color with smaller number chosen, $(M-k)$ balls of the other color is chosen. Order the balls, account for the identical objects.

• Can you explain it? Jul 18 '18 at 9:40
• @RafaelGonzalezLopez my bad just now, misread the question Jul 18 '18 at 9:46

The generating function for all words without restrictions is

$$GF(x,y,t) = {1 \over 1 - xt - yt}$$

where the coefficient of $$x^ny^kt^m$$ encodes the number of word of length $$m$$ that contain $$n$$ letters W and $$k$$ letters B.

$$\sum_{0 \leqslant n}^{N} coeff(x^n)\ \sum_{0 \leqslant k}^{K} coeff(y^k)\ \sum_{m=M}coeff(t^m) \ GF(x,y,t)$$

for the given case, this "viewport" will extract the polynomials:

$$x^3+3x^2y+3xy^2+y^3$$

$$x^3+3x^2+3x+1$$

$$7$$

Oh dear me ! here is a nice one :

We have to partition in two subsets a string of length M $$x_1x_2...x_M$$

then the exponential generating function is :

$$(1 + x + {x^2 \over 2!} + \cdots + {x^K \over K!}).(1 + x + {x^2 \over 2!} + \cdots + {x^N \over N!})$$

Just take the coefficient of $$x^M \over M!$$