# A pokemon stadium consists of 2 cat pokemon, 4 flying pokemon, 5 water pokemon, 12 fire pokemon and one psychic pokemon

How many distinguishable arrangements can be made if:

a) The 2 cat pokemon must come first and the psychic pokemon last?

• What I tried: 24P2*22P4*18P5*13P12*1P1 (This didn't give me the right answer)

b) The 2 cat pokemon must come first and the psychic pokemon last, however 4 flying pokemon have to be beside each other?

• Some hints: if a group of size $n$ must be at the front/back, then perform your calculations without them and then consider adding them back, recalling that there are multiple ways to arrange these $n$. If a group of size $n$ must be together in the permutation, consider them as a single element and perform your calculations, remembering to multiply the result by the number of distinct ways you could arrange that group of size $n$. – A.E Jul 7 '13 at 20:39
• I'm pretty sure, that cat is not a valid pokemon type, so the answer should be $0$. – Tomas Jul 7 '13 at 20:42
• @ Tomas: Meowth Is a cat type pokemon LOL. – MethodManX Jul 7 '13 at 20:43
• The shape of Meowth is a cat, the type however is Normal, as you can see here. However, we should move this discussion to pokemon.SE. – Tomas Jul 7 '13 at 20:46
• Okay guy I just need the answer for the question – MethodManX Jul 7 '13 at 20:47

Are pokemons of the same type distinguishable? If not, then for the first question the solution is as follows:

• total number of positions to fill equals 2 + 4 + 5 + 12 + 1 = 24

• you have the unique way to fill in the first two positions - by the two cat pokemon, and their order doesn't matter.

• you have the unique way to fill in the last position, that is, by the psychic pokemon.

• for the other 24 - 2 - 1 = 21 positions you have $A = \frac{21!}{4!\cdot 5! \cdot 12!}$ ways.

So there are $1\cdot A \cdot 1 = A$ ways.

The second question goes as follows. Since your four flying pokemon must be together (and are indistinguishable, hence it doesn't matter how they are arranged with respect to each other), you may consider them as a single object. So now you have not 24 objects and 24 positions, but in fact 21 object and 21 position. You deal with 2 cat pokemon and the psych pokemon as in the previous case, and for the rest 17 pokemon there are $\frac{17!}{5!\cdot 12!}$ different arrangements. Hope that helps...

• I get error when I try to compute that.. – MethodManX Jul 7 '13 at 20:55
• @MethodManX: It seems like you know the correct answer. Maybe you should tell us that, so that we can conclude, whether pokemon of the same type are distinguishable for the purpose of this task. – Tomas Jul 7 '13 at 21:00
• The answer is 37035180 for a – MethodManX Jul 7 '13 at 21:04
• @Tomas that's the answer – MethodManX Jul 7 '13 at 21:16
• @MethodManX: $\frac{21!}{4!5!12!}=37035180$, so Alex's answer is fine. (+1) – Tomas Jul 7 '13 at 21:50