Given 'a' identical objects of one kind and 'b' identical objects of other kind. Also, given 'k' indistinguishable buckets. In how many ways can one put the '(a+b)' objects into the 'k' buckets such that every bucket has atleast a single object?

As an example, let's suppose we have 3 As and 2 Bs and we need to partition them into 2 buckets. (a=3, b=2, k=2). The possible combinations are:

  1. A | AABB
  2. AA | ABB
  3. AAA | BB
  4. AAAB | B
  5. AAB | AB

So, there exist 5 such partitions.

  • $\begingroup$ Does the ordering withing the bucket matter (i.e. order in which the objects are put in the bucket)? $\endgroup$ – Aryabhata Apr 27 '12 at 21:41
  • $\begingroup$ No, the ordering within the buckets doesn't matter. $\endgroup$ – nitzs Apr 27 '12 at 21:46
  • $\begingroup$ I just saw your example. You say the buckets are different (distinguishable), but your example does not seem to make it so. $\endgroup$ – Aryabhata Apr 27 '12 at 23:23
  • $\begingroup$ Sorry, I didn't meant different as in distinguishable. The buckets are indistinguishable. $\endgroup$ – nitzs Apr 28 '12 at 7:30

It would be surprising if a closed form could be given for this number, since setting $b=0$ would give the number of partitions of $a$ into $k$ parts, for which no closed form is known. But we can readily write down a generating function by analogy with the partition number generating function: The desired number is the coefficient of $x^ay^bz^k$ in

$$\prod_{{\scriptstyle l,m=0}\atop{\scriptstyle l+m\ne 0}}^\infty\frac1{1-x^ly^mz}\;.$$

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Arrange in the a + b objects in a line. One can get $\frac{(a + b)!}{a! b!}$ such arrangements. Then, by the bars and stars theorems, we know that we can partition the arrangement of a+b objects into k buckets in ${a+b-1\choose{k-1}}$ possible ways. Therefore, in total,$$\frac{(a + b)!}{a! b!}{a+b-1\choose{k-1}}$$

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  • $\begingroup$ Won't this double count? $a| a bb$ and $a|b a b$ will be counted differently? (Assuming a bucket is a "multiset"). $\endgroup$ – Aryabhata Apr 27 '12 at 21:41
  • $\begingroup$ @Aryabhata Good point! I guess I will have to sit on it a bit more. $\endgroup$ – TenaliRaman Apr 27 '12 at 21:44

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