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Find the number of positive integer solutions to $a+b+c+d+e+f= 20$ subject to $1\leq a,b,c,d,e,f\leq 4$.

When there is only the lower bound, i.e $1\leq a,b,c,d,e,f$, we can substitute $g= a-1$, $h= b-1$, $i= c-1$, $j= d-1$, $k= e-1$, $l= f-1$ and then use the ball and sticks method to find the number of solutions to the equation $g+h+i+j+k+l= 14$. Similarly when there is only the upper bound, i.e. $a,b,c,d,e,f\leq 4$, we can substitute $g= 4-a$, $h= 4-b$, $i= 4-c$, $j= 4-d$, $k= 4-e$, $l= 4-f$ and then use the ball and sticks method to find the number of non negative integral solutions to the equation $g+h+i+j+k+l= 4$. But what will be the approach when there is both upper bound and lower bound? When there was only the upper bound we knew that $g,h,i,j,k,l$ are always non negative, but they can be any non negative integers, because $a,b,c,d,e,f$ can be negative which implies $a,b,c,d,e,f$ can also be greater than $4$. Now when two bounds are introduced, we still know that $g,h,i,j,k,l$ are non negative but they cannot be greater than $4$. So we cannot use the ball and sticks method that way. My question is how do we approach in such cases?

abcd
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1 Answers1

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You could use the generating function method: the number of solutions with sum $n$ is the coefficient of $x^n$ in $(x + x^2 + x^3 + x^4)^6$.

Robert Israel
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  • That method too works, so does the ball and bars method. But what will be the case when the variables are subject to an upper and lower bound? That is what I asked in my question. – abcd May 31 '13 at 15:06
  • I did use the lower bound ($1$). That's why it's $x+\ldots$ instead of $1+x+\ldots$. – Robert Israel May 31 '13 at 15:07