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How many possibilities are there:

  • neither dolls nor girls differ from each other and every girl gets at least one doll? Is the answer $\binom{n-1}{k-1} = \binom{41}{22}$ ???

The idea is to draw it in this way. o_o_o_o_o_ ... _o

Where o stands for doll and _ stands for the possible separating line between two sets. And because we have 41 underlines and we have to put 22 lines in order to separate the groups we take binomial coefficient of (41 22)

  • only dolls differ from each other
  • only girls differ from each other. Is the answer $\binom{n+k-1}{k-1} = \binom{42+23-1}{23-1} = \binom{64}{22}$ ???
  • both dolls and girls differ from each other

Does it mean that every doll can be assigned to any of the girls. So it means that for every doll we have 42 possibilities. So $23^{42}$ ???

  • both dolls and girls differ form each other and every girl has at least one doll
  • neither dolls nor girls differ from each other
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    If each girl gets at least one doll, and the dolls are indistinguishable, then that's the same as giving each girl a doll, and then having $19$ dolls left to distribute without any restrictions. If the girls are not different from one another, then the number of ways to do this is called the _number of partitions of $19$_, and there is no simple formula for it. You just have to count. – 2017-01-17
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    @ThomasAndrews sorry, I have forgotten to write down my answer – 2017-01-17
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    @Arthur That's only true when the dolls are all the same, though. – 2017-01-17
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    The answer $\binom{41}{22}$ is the answer for "only girls differ from each other." – 2017-01-17
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    @ThomasAndrews ",,, and each girl gets at least one doll." – 2017-01-17
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    Oh, right, I was thinking that was a given condition for all the questions, but I guess not. @Joffan – 2017-01-17
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    @ThomasAndrews no, it wasn't given for the whole exercises, only for one question – 2017-01-17

3 Answers 3

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I'm only going to deal with the first bullet.

Give each girl one doll. The remaining 19 dolls can be partitioned into nonempty heaps in $490$ ways. There is no simple combinatorial formula for these partition numbers , but a famous recursion attributed to Euler.

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Neither Dolls nor Girls Differ from Each Other and Each Girl is Given at Least One Doll

In this answer, it is shown that given $m$ indistinguishable bins, the generating function of the number of ways to fill the bins with $n$ indistinguishable items is $$ \prod_{k=1}^m\frac1{1-x^k} $$ In this case, $m=23$ and, after giving one doll to each girl, we have $19$ dolls to be distributed: $$ \begin{align} \prod_{k=1}^{23}\frac1{1-x^k} &=1+x+2x^2+3x^3+5x^4+7x^5+11x^6+15x^7+22x^8\\ &+30x^9+42x^{10}+56x^{11}+77x^{12}+101x^{13}+135x^{14}+176x^{15}\\[9pt] &+231x^{16}+297x^{17}+385x^{18}+490x^{19}+O\!\left(x^{20}\right) \end{align} $$ Thus, there are $490$ arrangements.

Only Girls Differ from Each Other

There are $42$ dolls to distribute among $23$ girls. This can be done using the "stars and bars" method.

Here is a simple example with three dolls among five girls: $$ \star\mid\mid\star\mid\star\mid $$ There are $4$ bars separating areas for the $5$ girls. There are $3$ stars representing the $3$ dolls. The number of arrangements is thus, $\binom{7}{3}$, where $7$ is the number of dolls plus the number if girls minus one (stars plus bars minus one). and $3$ is the number of dolls (stars).

In the problem at hand, this gives $$ \binom{42+23-1}{42}=\binom{64}{42}=\binom{64}{22} $$

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    Except that if the girls don't differ, then stars and bars doesn't work, and we are looking for the number of (unordered) partitions of $19$ into 22 numbers. – 2017-01-17
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    There are six bullets in the question. To which one are you referring? – 2017-01-17
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    @ChristianBlatter: when I first read the question, there were no bullets. Now the question is much different. I have answered two of the bullets and noted which is which. – 2017-01-17
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Girls and dolls differ

There are $23^{42}$ ways to give the dolls to girls. You can give each doll to any of the $23$ girls.

You are going to need to an inclusion-exclusion argument for the case where every girl gets at least one doll. Letting $A_i$ be the ways to give the $42$ dolls to $23$ girls where girl $i$ does not get any dolls. Then you want to count:

$$23^{42}-|A_1\cup A_2\cdots\cup A_{23}|$$

Inclusion-exclusion gives you:

$$\sum_{k=0}^{22}(-1)^k\binom{23}{k}(23-k)^{42}$$

Girls differ, dolls don't

Then you are correct that it is $\binom{42+23-1}{23-1}$, and it is $\binom{42-1}{23-1}$ if each girl gets at least one.

For the other cases, you get into the messy area of partitions of sets and of numbers.

For example, if both girls and dolls are indistinguishable, you want the number of ways to write $42=a_1+a_2+\cdots+a_{23}$ with $0\leq a_1\leq a_2\leq \cdots\leq a_{23}$. And if every girl gets at least one doll, then you are counting only the cases $1\leq a_1$.

Partitions don't have remotely simple counting formulas. (If you watch the movie "The Man Who Knew Infinity," a big part of the second half of that movie is Ramanujan and Hardy working on some identities related to partition formulas.)

You can represent the case when only dolls are different in terms of Stirling numbers of the second kind:

$$\sum_{k=1}^{23}\left\{42\atop k\right\}$$

and, if every girl gets at least one doll:

$$\left\{42\atop{23}\right\}$$

You actually have the "obvious" relationship:

$$\left\{42\atop{23}\right\}=\frac{1}{23!}\sum_{k=0}^{22}(-1)^k\binom{23}{k}(23-k)^{42}$$

where the complex sum is the same value for the girls distinct.