Division of n Identical Objects into r identical groups I know that the formula $C(n-1,r-1)$ is used when we need to distribute n identical objects in $r$ distinct groups(such that every group receives atleast 1 object).
Suppose 10 objects need to be divided into 3 groups such that each group receives atleast one object. By using the above formula , we also consider the case such as $(1,2,7)$ $(1,7,2)$ $(2,1,7)$ $(2,7,1)$ $(7,1,2)$ $(7,2,1)$ ways.
But in this case we need to consider only one out of these possibilities.
One might think of using $C(n-1,r-1)$ and then dividing the result by $r!$. That way you should get the answer as $C(9,2)\div3! = 6$. But by writing out all the possibilities,  you get 8 different ways!
So is there any formula/algorithm to solve this?.
P.S: I have tried the following (for the above question)

*

*Consider first digit to be 1 . Therefore The division is of the form $(1,x_1,x_2)$. Now we apply the above formula for solutions of $x_1$ and $x_2$ and divide it by $2!$


*Consider first digit to be 2. Therefore the division is of the form $2,y_1,y_2$ such that $y_1>2$ and $y_2>2$ (Since the cases where $y_1$ and $y_2$ is less than or equal to 2 are already considered in step 1). We do some simple shifting of values and find the ways in which this can be done


*Similarly we consider first digit to 3 and continue in the above manner and calculate the number of ways.


*In the end add out the possibilities of all the cases
While this algorithm works for division into 3 groups , but it will become very long if the division is extended more.
So is there any generalised formula for doing this?
 A: You can use this method. Your question is
How to distribute n identical things into r identical groups?
Suppose you have a 10 identical objects and 3 identical groups. If you were given that the sizes of the group are in the order $g_{1}\geq g_{2}\geq g_{3}$ , you would have only considered $(7,2,1)$. Applying this for a general case.
Algorithm: To distribute n identical objects into r(here, r=3) similar groups of differing(some maybe equal) sizes:
Let the group sizes be denoted by $x_{1},x_{2},x_{3}$ then $x_{1}+x_{2}+x_{3}=n$
Also, since $x_{1}\geq x_{2} \geq x_{3}$, let $$x_{1}=x_{3}+a+b$$
$$x_{2}=x_{3}+a$$
$$x_{3}=x_{3}$$
Adding $x_{1},x_{2},x_{3}$,
$3(x_{3}+1)+2a+b=n$ (where a,b,$x_3$ are whole numbers)
or, $3x_{3}+2a+b=n-3$
Finding the number of distinct tuples for (a,b,$x_3$) will give you the answer.
The tuples set for $n=10$ is
$\{(2,0,1),(1,0,4),(1,1,2),(1,2,0),(0,0,7),(0,1,5),(0,2,3),(0,3,1)\}$
Algorithm: To distribute n identical objects into r similar groups of differing(some maybe equal) sizes:
Let the group sizes be denoted by $x_{1},x_{2},x_{3}...x_r$ then $x_{1}+x_{2}+x_{3}...x_{r}=n$
Also, since $x_{1}\geq x_{2} \geq x_{3}...\geq x_{r}$,
let $$x_{r}=x_{r}$$
$$x_{r-1}=x_{r}+c_1$$
$$...$$
$$x_{1}=x_{r}+c_1+c_2+...c_r$$
Adding $x_{1},x_{2},x_{3},...x{r}$,
$r(x_{r}+1)...+2c_2+c_1=n$ (where $c_1,c_2,...,x_r$ are whole numbers)
or, $$rx_{r}...+2c_2+c_1=n-r$$ (where $c_1,c_2,...,x_r$ are whole numbers)
Finding the number of distinct tuples for ($c_1,c_2,...,x_r$) will give you the answer. This can be done using multinomial-theorem or a computer program if the number n and groups are large.
A follow up for the answer has been asked and can be linked here: How to use Bars and Stars method for equations with more than 1 non unity coefficients?
