Prove the following theorem of distributing identical objects

The number of ways to distribute $r$ $identical$ $objects$ into $n$ $distinct$ $boxes$ with at least $r_i$ objects in the $i^{th}$ box is:

$$C\left(\left(r-r_1-r_2-...-r_n\right)+n-1,n-1\right)$$

For example, suppose I had 12 identical objects in 4 distinct boxes, and with at least 1 object in $n$ boxes (each one).

Then I get:

$$C\left(\left(r-n\right)+n-1,n-1\right)$$

How can I prove the general formula listed above?

• Load the required items, distribute the rest with stars and bars. – Joffan Dec 11 '16 at 4:21
• I need to prove it, I'm not sure reasoning with logic would work. What would be a more mathematical approach? – K Split X Dec 11 '16 at 4:23
• Obviously that was just a sketch, but some degree of logical reasoning is generally required to hold the mathematics together. – Joffan Dec 11 '16 at 4:40

Let the $i^{th}$ box contain $x_{i}$ objects. Then the following holds: $$x_1 + x_2 + x_3 + ..... + x_n = r$$ Now due to the constraint mentioned $x_i \ge r_i$. So, do the following substitution: $$x_i = x_i^{'} + r_i$$ Hence all $x_i^{'} \ge 0$, i.e. they are all non-negative.
Now, the equation becomes, $$(x_1^{'} + r_1) + (x_2^{'}+r_2) + ..... +(x_n^{'}+r_n) = r \implies$$ $$x_1^{'}+x_2^{'}+x_3^{'}+.....+x_n^{'} = r - (r_1+r_2+r_3+.....+r_n)$$
Now since all $x_i^{'}$ are non-negative, the Multinomial Theorem can be used to calculate the non-negative solutions of the above mentioned equation(which are in-turn equal to the solutions of the original equation before substitution).
And after using Multinomial theorem, you get the number of ways as $\binom{(r-r_1-r_2-...-r_n)+n-1}{n-1}$.