The students $S_1, S_2,...S_{10}$ are divided into 3 groups A, B and C

Question

It can be seen as distributing n unique objects above m groups

The students $S_1, S_2,...S_{10}$ are divided into 3 groups A, B and C such that each group has at least one students and C has at most 3. Find possibilities of forming groups

$$a+b+c=10$$

The number of positive integral solutions $(\ge 1)$ is $$\binom {10-1}{3-1}=36$$

I don’t know how to account for the cases where C has at most 3 students

Also I know this method of solving is probably wrong but I don’t know how else to do it.

@user most likely $atmost$ – Albus Dumbledore Mar 02 '21 at 16:24 — Albus Dumbledore, Mar 02 '21 at 16:24

true blue anil · Accepted Answer · 2021-03-02T18:19:55.197

Leaving aside group $C$, each student has $2$ choices for groups

Group $C$ can receive a student in $\binom{10}1$ ways.

The remaining students could be received by the other two groups in $2^9$ ways, but this could result in one group receiving no student.

Any of the two groups could be empty, so we subtract $2$ to get legal ways as

$\binom{10}1\cdot (2^9-2)$, and proceeding similarly with group $C$ having $2$ or $3$ students, and adding up, we get

$\binom{10}1\cdot (2^9-2) + \binom{10}2\cdot(2^8-2) +\binom{10}3\cdot(2^7-2) = 31650$

user · Answer 2 · 2021-03-02T17:04:35.700

1

Hint: the simplest way is to consider the problem by cases. There are only 3 options: $$ c=1,2,3. $$

The second issue is that you are asked not about the number of ways to choose the size of groups but to distribute the students between them. For example for $c=1$ the number is $$ \binom{10}1\sum_{i=1}^8\binom9i=\binom{10}1(2^9-2) $$

Can you take it from here?

edited Mar 02 '21 at 17:04

answered Mar 02 '21 at 16:29

user

26,272

user--- Are you saying it's easy as 1,2,3? :-) – coffeemath Mar 02 '21 at 16:31
@coffeemath I am saying $c=1$ or $c=2$ or $c=3$. :) – user Mar 02 '21 at 16:32
Yes maybe I shouldn't have left a comment which is only a jest... Don't want to confuse the question asker... – coffeemath Mar 02 '21 at 16:36
I think your comment was very useful for clarifying the issue. – user Mar 02 '21 at 16:38
Ok, so then what would give an equation with a and b, after which we just use the method I used above? – Aditya Mar 02 '21 at 16:45
That will give $8+7+6=21$ but the correct answer is 31650 – Aditya Mar 02 '21 at 16:55
@Aditya I have added the next step. Hope it helps. – user Mar 02 '21 at 17:05
I don’t understand how you got that. What are you using exactly? – Aditya Mar 02 '21 at 17:10
@Aditya I was using $\sum_{i=0}^n\binom ni=2^n$. – user Mar 02 '21 at 17:13
No I meant theoretically, what did you use? Like how did you get the expression with the coefficients? – Aditya Mar 02 '21 at 17:21
1

I count the numbers of ways to choose a student for the group $C $ and then those to choose $i $ students (from the 9 remaining) for the group A. The rest goes to the group B. – user Mar 02 '21 at 17:35

The students $S_1, S_2,...S_{10}$ are divided into 3 groups A, B and C

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