The problem reads:
"I have 15 certificates for a free pizza and 24 cans of Coca-Cola. How many ways may I distribute the certificates and the cans of coke to 22 students?"
The answer is given as: 36 C 21 * 45 C 21
I have trouble understanding where 21 came from if we are given 22 students in condition of the problem.
$\endgroup$ 21 Answer
$\begingroup$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 $C_{36}^{21}C_{45}^{21}$.
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