# Sampling with replacement order doesn't matter problem

At first you need distribute pizze to 22 students. Assume stars $*$ represent the pizza and each student like a box $\\|$ represent an end side of a box. $$\underbrace{*\ *\ *\ \ \ \ \ \ *}_{15\ balls}\ \ \ \underbrace{[ \ \vert \ \vert \ \vert \ \vert \ \ \ \ \vert \ \vert \ ]}_{22 \ boxes}^{21\ bars}$$ Take two of the bars as special, to represent left and right ends. Then the original problem may be reformulated : How many different combinations of these $15+22-1$ objects there are? This is $${(15+22-1)!\over 15!\cdot (21)!} = \binom{36}{21}=C_{36}^{21}$$ For Coca-cola it is ${(24+22-1)!\over 24!\cdot (21)!} = \binom{45}{21}=C_{45}^{21}$. And the common variants is the product $C_{36}^{21}C_{45}^{21}$.