Say I have Group A and Group B

Group A needs 1 student and group B needs 2 students. There are 3 students total (A,B,C).

What sort of formula could I use to determine the total number of assignment combinations of students to groups.

So for the above example I could have

Group 1   Group 2
A         BC
B         AC
C         AB

Or if I had 3 groups, 3 students and each group needs 1 student

Group 1   Group 2   Group 3
A         B         C
A         C         B
B         A         C
B         C         A
C         A         B
C         B         A

Obviously I'm going to use this on a much larger scale, I've been reading up on combinatorics but I haven't been able to find a formula to use for this that could cover both cases.



Assuming you designate all the $n$ students to groups of predesignated group sizes $\{g_i\}$ (so that $\sum g_i = n$), you have $$\frac{n!}{\Pi (g_i!)}$$

In the first case, $n=3, \{g_i\} = \{1,2\}$ and there are $\frac{3!}{1!2!}=3$ choices

In the second case, $n=3, \{g_i\} = \{1,1,1\}$ and there are $\frac{3!}{1!1!1!}=6$ choices

  • $\begingroup$ Thank you, that is exactly what I was looking for. $\endgroup$ – Belgin Fish Jan 21 '15 at 22:50
  • 1
    $\begingroup$ As you can see, in the case of two classes, effectively you are just choosing the students for one class (and the other class is "the rest") and the value is the normal choose-k-from-n. $\endgroup$ – Joffan Jan 21 '15 at 22:53

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