Number of ways of distributing $4$ identical red balls,$1$ green ball,$1$ black ball among $4$ persons The question is to find out the number of ways of distributing :


*

*$4$ identical red balls

*$1$ white ball

*$1$ green ball

*$1$ black ball 


among $4$ persons if each receives at least one ball and no one gets all identical red balls.

My attempt

Let $a_i$ denote the number of balls the $i^{th}$ person has. So we want the solution to the equation $$\sum a_i=7\quad\text{ where } 0<a_i<5$$ The total number of  integral solution to this equation is $20$.
Since all balls are not identical the total number of ways become $20*\frac{7!}{4!}$ which is $4200$.
The number of ways in which a person gets all identical red balls is $\binom 41 3!$ we have to subtract this value. So the number of ways sought in the question should be $4200-24=4176$. 
However this answer is incorrect. Can anyone explain where I went wrong.
Thanks.
 A: In the following, I call $A,B,C,D$ respectively the first, second, third and last player to get at least one stone. So the letters are not referring to a specific player. Instead this notation focuses on making clear how many players have been served so far during ball distribution.



*
  
*giving $1,1,1,1$ red balls : $1$ choice
  

$4$ people are served so we are free to distribute remaining balls


  
*  
    
*giving green ball : $4$ choices
    
*
      
*giving white ball : $4$ choices
      
*
        
*giving black ball : $4$ choices
      
    
  


$\text{Subtotal}=1\times(4\times 4\times 4)=64$




*
  
* giving $1,1,2$ red balls : $\frac{4\times 3}{2}\times 2=12$ choices (we divide by $2$ because $1,1$ has no order)
  

$3$ people $A,B,C$ are now served

 
  
*
    
* giving green ball to $A,B,C$ : $3$ choices
    
*
      
* giving white ball to $A,B,C$ : $3$ choices
      
*
        
* giving black ball to $D$ : $1$ choice
      
      
* giving white ball to $D$ : $1$ choice
      
*
        
* giving black ball : $4$ choices
      
     
    
* giving green ball to $D$ : $1$ choice
    
*
      
* giving white ball : $4$ choices
      
*
        
* giving black ball : $4$ choices
      
    
  


$\text{Subtotal}=12\times\bigg(3\times\big((3\times 1)+(1\times 4)\big)+1\times(4\times 4)\bigg)=12\times(3\times7+16)=444$




*
  
*giving $2,2$ red balls : ${4 \choose 2}=6$ choices
  
*giving $1,3$ red balls : $4\times 3=12$ choices
  

In both cases only $2$ people $A,B$ are served.


  
*
    
*giving green ball to $A,B$ : $2$ choices
    
*
      
*giving white ball to $C$ : $2$ choices
      
*
        
*giving black ball to $D$ : $1$ choice
      
    
  
  
*
    
*giving green ball to $C$ : $2$ choices
    
*
      
*giving white ball to $A,B,C$ : $3$ choices
      
*
        
*giving black ball to $D$ : $1$ choice
      
    
    
*
      
*giving white ball to $D$ : $1$ choice
      
*
        
*giving black ball : $4$ choices
      
    
  
   

$\text{Subtotal}=(6+12)\times\bigg( 2\times (2\times 1)+2\times\big((3\times 1)+(1\times 4)\big)\bigg)=18\times(4+2\times7)=324$

$\text{Total}=64 + 444 + 324 = 832$

