A question on combinatorics: if $p + q + r = 10$, find the number of possible combinations where the condition holds true.

Question

If $ p + q + r = 10 $, find the number of possible combinations satisfying the equation. ( $p,q,r ≥ 0$)

---

PS - As suggested by Brian Moehring, I used the 'stars and bars' method.

We need the sum as $10$. We have to add $1$ ten times, to get 10. If we add two bars i.e., we are dividing them into p, q and r. So now in total there are $10 \ 1's$ and $2$ bars. We have to find the number of ways to arrange those $2$ bars in $12$ places. Hence, there are $\binom{12}{2}$ ways of selecting $p,\ q$ and $r$ to satisfy the equation.

Answer 1

The Stars and Bars Theorem Provide this formula that we can use here:

$$\binom {n+k-1}{n}=\binom {n+k-1}{k-1}$$

Using $n=10,k=3$

$$\binom {10+3-1}{3-1}=\binom {12}{2}=66$$

Answer 2

Let do this: $$p+q+r=10 \Longrightarrow (p+1)+(q+1)+(r+1)=13$$ $$\left\{\begin{array}{c}p'=p+1>0 \\ q'=q+1>0 \\ r'=r+1>0 \\ \end{array}\right\}\Longrightarrow p'+q'+r'=1+...+1$$ That in right side of last equation we have $13$ ones. Since new variables are positive, any solution is equivalence to choose $2$ number of $+$ in right side as candidate for that we have at left side between variables. So the number of solution is: $$\binom{12}{2}=66$$