How many ways there are to group 27 same cookies into 3 same groups, where each group contains at least 1?

If the groups are distinct, then I can use the technique to calculate how many vectors(x1,x2,x3) there are such that x1 + x2 + x3 = 27. However, the groups are actually same, how can I solve that? Thanks.

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shouldnt we change group to set? –  Bananarama Jan 8 '13 at 3:13

You want the number of partitions of $27$ into at most $3$ parts; this is given by the partition function $q(n,k)$, which gives the number of partitions of $n$ into at most $k$ parts. Specifically, you want $q(27,3)$. The partition function satisfies a simple recurrence, so it’s possible to calculate small values fairly easily. However, the numbers $q(n,3)$ are OEIS A001399, where you can read off the value $75$. You can also find the formula

$$q(n,3)=\left\lfloor\frac{(n+3)^2}{12}+\frac12\right\rfloor\;.$$

Added: The recurrence $$q(n,k)=q(n,k-1)+q(n-k,k)$$ can be justified as follows. Partitions of $n$ into at most $k$ parts are of two kinds: those that have fewer than $k$ parts, and those that have exactly $k$ parts. There are $q(n,k-1)$ that have fewer than $k$ parts. Now take a partition with exactly $k$ parts; if you remove $1$ from each part, you have a partition of $n-k$ into at most $k$ parts. Conversely, if you start with a partition of $n-k$ into at most $k$ parts and add $1$ to each part, you get a partition of $n-k$ into exactly $k$ parts. Since there are $q(n-k,k)$ partitions of $n-k$ into at most $k$ parts, there must be $q(n-k,k)$ partitions of $n$ into exactly $k$ parts. And this justifies the recurrence: the $q(n,k)$ partitions of $n$ into at most $k$ parts include $q(n,k-1)$ that have fewer than $k$ parts and $q(n-k,k)$ with exactly $k$ parts.

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what about exact 3 groups? that being said, each group has at least 1. –  Frank Xu Jan 8 '13 at 0:02
@Frank: If each group has at least $1$, then you are in effect distributing $24$ arbitrarily, so just use $q(24,3)=61$ instead. –  Brian M. Scott Jan 8 '13 at 0:04
could u explain more about q(n,k) = q(n,k-1) + q(n-k,k)? I don't quite understand the q(n-k,k) part, thank you –  Frank Xu Jan 8 '13 at 0:26
@Frank: It comes from the same reasoning as my previous comment; I’ve expanded on that in an addition to my answer. –  Brian M. Scott Jan 8 '13 at 0:34

The size $x_1$ of the largest group ranges from $9$ to $25$. From $x_1=9$ up to $x_1=13$ there are $1,2,4,5,7$ options, respectively, for splitting up the rest. From $x_1=14$ up to $x_1=25$ there are $6,6,5,5,4,4,3,3,2,2,1,1$ options, respectively, for splitting up the rest. Thus the total is $1+2+4+5+7+2\cdot6(6+1)/2=19+42=61$, in agreement with Brian's more elegant answer.

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