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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.

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  • $\begingroup$ I have no reputation for comment but this topic math.stackexchange.com/questions/1441170/… $\endgroup$
    – user518463
    Commented Jan 4, 2018 at 9:37
  • $\begingroup$ Thank you for referral! However, I need help with this particular question. $\endgroup$
    – Stacy
    Commented Jan 4, 2018 at 9:41

1 Answer 1

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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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  • $\begingroup$ Thank you! I think I got it! $\endgroup$
    – Stacy
    Commented Jan 4, 2018 at 9:59

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