1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $C_{36}^{21}C_{45}^{21}$.

$\endgroup$
  • $\begingroup$ Thank you! I think I got it! $\endgroup$ – Stacy Jan 4 '18 at 9:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.