Why the result is different in combinatory exercise?

Question

I have a statement that says:

In a bingo, each card can be used for two games, in how many ways Can you distribute the 2 biggest prizes, if there are 10 cards in total?

My development was:

i have 2 prizes, 10 cards and each card can be used for two games.

When the first prize is won, i have $\frac{(n + r - 1) !}{(n - 1)! * r !} = \frac{(10 + 1 - 1) !}{(10 - 1)! * 1 !}$ = $\frac{10!}{9!} = 10$ possible winners, and when the second prize is won, i have $10$ too.

And according to the additive principle, is $20$, since the games do not happen simultaneous.

But my problem, is that the correct answer must be $100$, so I want to know, what have I done wrong?

Answer 1

Your mistake is in using the additive principle. We use the additive principle when we count disjoint possibilities, so you're saying there are $10$ ways for someone to get the first prize, and $10$ says for someone to get the second prize, and $20$ total ways for a single person to get a single prize. The results you're looking for are who gets the first prize and who gets the second prize, so each possibility distributes both prizes. Then you have to use the multiplicative principle, and get $10*10=100$ possibilities.

Think of it like clothing- if you have $3$ shirts and $4$ pairs of pants, taking the sum would count ways of picking a single shirt, or picking a single pair of pants. But if you want to count number of ways to pick a complete outfit, both shirt and pants, you have to take the product.

Answer 2

You have mistaken what the results are that you are adding.

The first set of results:

$\{\text{Card 1 wins the first game},\text{Card 2 wins the first game},\text{Card 3 wins the first game},\dots,\text{Card 10 wins the first game}\}$

The second set of results:

$\{\text{Card 1 wins the second game},\text{Card 2 wins the second game},\text{Card 3 wins the second game},\dots,\text{Card 10 wins the second game}\}$

Although the time at which the games occur is not "simultaneous" the results are both able to happen. It is possible for card1 to both win the first game and win the second game for example.

Compare this to a correct time at which you add such as $\{\text{Card 1 wins the first game},\text{Card 3 wins the first game},\text{Card 5 wins the first game},\text{Card 7 wins the first game},\text{Card 9 wins the first game}\}$ $\{\text{Card 2 wins the first game},\text{Card 4 wins the first game},\text{Card 6 wins the first game},\text{Card 8 wins the first game},\text{Card 10 wins the first game}\}$

giving $5+5$ ways in which the first game could be won since it is (presumably) impossible for both Card1 and Card2 to both win the first game as each game (presumably) only has one winner.

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That being said, if all you care about is who received the prizes, but you don't care about what times it occurred at (e.g. if the prize for game1 is the same as the prize for game2) then there would have been $\binom{10}{2}+10=55$ possible results, not $100$.