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An opaque bag contains $48$ distinctly numbered balls, equally split into $8$ colours (so $6$ balls each) and $35$ balls are extracted from it. In how many ways, $F_k(8,6,35)$, can we do so such that exactly $k$ colours are completely extracted?

Obviously, for $k \geq 6$ the answer is $F_k(8,6,35) = 0$ since we don't extract enough balls to make $k$ full sets. I have been given the formulae for $k=5$ and $k=1$. Namely, $$F_5(8,6,35) = \binom{8}{5}\binom{18}{5},$$ $$F_1(8,6,35) = \binom{8}{1}\left[\binom{7}{0}\binom{42}{29} - \binom{7}{1}\binom{36}{23} + \binom{7}{2}\binom{30}{17} - \binom{7}{3}\binom{24}{11} + \binom{7}{4}\binom{18}{5}\right].$$

I think my interpetation for $F_5$ is correct. It essentially boils down to selecting which of the $5$ colours out of $8$ to complete, and then extracting $35 - 6\cdot 5 = 5$ balls out of the remaining $48 - 6\cdot 5 = 18$ ones. Note that since we are left with extracting fewer balls than necessary to complete another full set, no double-counting occurs.

For $F_1$ I am stumped. It seems to involve some sort of inclusion/exclusion principle but I cannot interpret any of the terms except for the first one, where we select which out of the $8$ colours to complete and then select $35 - 1\cdot 6 = 29$ balls left to extract out of the remaining $48 - 1\cdot 6 = 42$ balls. Clearly, since we can end up extracting complete sets in those $29$ out of $42$ balls, we are double-counting in some way.

My questions are:

  1. What is the interpretation of the formula for $F_1$?
  2. What are the formulae for the other $F_k$?
  3. (Generalisation) If we have $c\cdot b$ distinctly numbered balls in an opaque bag split into $b$ balls of each of $c$ colours, in how many ways, $F_k(c,b,e)$, can we extract $e$ balls so that exactly $k$ colours are completely extracted?
Darth Geek
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1 Answers1

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Your reasoning for $F_5(8, 6, 35)$ is correct.

Here is how the reasoning for $F_1(8,6,35)$ goes. There are $\binom81$ ways to choose the completely extracted color; suppose from now on the extracted color is fixed, say it is red. We now apply the principle of inclusion-exclusion to count the number of ways to extract $29$ additional balls from the remaining $42$ non-red balls, such that no other color is completely extracted.

There are then $\binom{42}{29}$ ways to choose the remaining balls; this is the size of our "universe" for PIE. For each of the $7$ other colors, we need to subtract out the $\binom{36}{23}$ distributions in our universe where both that color and red were completely extracted. There are $\binom71$ ways to choose the other color, so we subtract $\binom 71\cdot \binom{36}{23}$.

In the previous step, distributions which completely extracted red and two other colors were double subtracted. These need to be added back in to correct for double subtraction. There are $\binom 72$ pairs of non-red colors, and for each pair, there are $\binom{30}{17}$ ways to ensure that red and both of those non-red colors were fully extracted, so we add in $\binom 72\cdot \binom{30}{17}$ for the next step of the principle of inclusion exclusion. The remaining steps all proceed analogously.

Applying the same reasoning, you can show in general that

$$ F_k(c,b,e)= \binom{c}k\cdot \sum_{i\ge 0}(-1)^i \binom{c-k}{i}\binom{bc-b(k+i)}{e-b(k+i)}\tag1 $$ Even though the sum ranges over all natural numbers $i$, the sum is effectively finite, because any term for which $e-(b(k+i))$ is negative will be zero, as $\binom nk$ is zero when $k$ is negative.

Here is a different method, using the generalized principle of inclusion exclusion. Let $S$ be the set of all $\binom{bc}{e}$ ways to choose $e$ balls from the urn. For each $i\in \{1,\dots,c\}$, let $E_i$ be the set of ways where the $i^\text{th}$ color is completely chosen. Then the number of ways to completely extract exactly $k$ colors is $$ \begin{align} \#\{\text{distributions in exactly $k$ of $E_1,\dots,E_c$}\} &= \sum_{j=k}^{c}(-1)^{j-k}\binom jk\sum_{1\le i_1<\dots<i_j\le b }|E_{i_1}\cap \dots \cap E_{i_j}| \\&= \sum_{j=k}^{c}(-1)^{j-k}\binom jk\binom{c}{j}|E_1\cap \dots \cap E_j| \\&= \sum_{j=k}^{c}(-1)^{j-k}\binom jk\binom{c}{j}\binom{bc-bj}{e-bj} \\&= \sum_{j=k}^{c}(-1)^{j-k}\binom ck\binom{c-k}{j-k}\binom{bc-bj}{e-bj}\tag{2} \end{align} $$ You can check that $(1)$ and $(2)$ are equal to each other, which can be seen by reindexing the sum in $(1)$ using $i\gets j-k$.

Mike Earnest
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