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There is a contest with 40 Pokemons. There are 18 Pokemons who like to fight in the sky, and 23 who like to fight on ground. Several of them like to fight in water. The number of those who like to fight in the sky and on ground is 9. There are 7 Pokemons who like to fight in the sky and in water, and 12 who like to fight on ground and in water. There are 4 Pokemons who like to fight in the sky, on ground, and in water. How many Pokemons like to fight in water

How do i express this problem with discrete math?

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Is there any pokemon who don't want to fight? –  Quixotic Mar 16 '12 at 7:53
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@Foool No Pokemon wants to fight, but Trainers can be so cruel... –  Austin Mohr Mar 16 '12 at 7:55
    
Venn diagram –  pedja Mar 16 '12 at 7:56
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S -set of pokemons who like to fight in the sky , W - set of pokemons who like to fight in water , G - set of pokemons who like to fight on the ground....Answer should be 23 . –  pedja Mar 16 '12 at 8:09
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Why the downvote? –  joriki Mar 16 '12 at 8:13

3 Answers 3

up vote 2 down vote accepted

Let' denote : S -set of pokemons who like to fight in the sky , W - set of pokemons who like to fight in water , G - set of pokemons who like to fight on the ground .

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$x=40-(4+3+5+8+6+6)=8$ , hence :

$P_W=3+4+8+8=23$

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Assuming there every Pokemon wants to fight.

We can use principle of inclusion and exclusion to solve this problem. If $x$ is the number of Pokemons like to fight in water then, $$40 = 18+23+x -(9+7+12) + 4 \Rightarrow x = 23$$

If there are Pokemon who don't want to fight then we need to subtract that number from the RHS of the equation and then solve for $x$.

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Why do you call it "mutual"? If I understand correctly this is the same as what's otherwise just called "inclusion-exclusion"? –  joriki Mar 16 '12 at 8:15

$N(sky\cup ground\cup water)=N(sky)+N(ground)+N(water)-N(sky\cap ground)-N(sky\cap water)-N(ground \cap water)+N(sky\cap ground\cap water)$

where N(X) is number of pokemon that wants to fight in terrain X. This is called inclusion-exclusion principle.

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