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There are 10 students. Find the number of ways in which they can be divided into 3 groups such that each group has at least 1 student and third group has at most 3

We can make 3 separate cases where $G_3$ has $1$, $2$, $3$ students respectively.

As for the remaining two groups, there are no restrictions other than that no group can have $0$ students.

But I am not sure about how to divide the remaining students into $2$ groups. Clearly the stars and bars method won’t work because students are not identical objects.

Arctic Char
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Aditya
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2 Answers2

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Let's take the case where $G_3$ has one student. Pick the student there and you have nine left. Now, $G_1$ (say) can have $1$ to $8$ students. Pick the students for $G_1$ and then all of the others are on $G_2$.

So for this case you'll get

$$10\cdot \sum_{j=1}^8 {9 \choose j}$$

groups. And similarly for the other cases for $G_3$.

John
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2

You can imagine this as labelling the students with the group numbers $\{1,2,3\}$.

So, basically you are looking for all $10$-digit numbers with digits $\{1,2,3\}$ where there is at least one $1$, one $2$ and from one to three $3$'s.

Now, splitting up into the cases where exactly $1,2,3$ get the label $3$, you get

  • exactly $1$ with label $3$: $\color{blue}{\binom{10}1\cdot (2^9-2)}$
  • exactly $2$ with label $3$: $\color{blue}{\binom{10}2\cdot (2^8-2)}$
  • exactly $3$ with label $3$: $\color{blue}{\binom{10}3\cdot (2^7-2)}$

Adding this together gives:

$$\color{blue}{31650}$$

trancelocation
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