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?


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*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?
Please explain how you got your answer I want to learn thanks:)
 A: Are pokemons of the same type distinguishable? If not, then for the first question the solution is as follows:


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*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...
