If we want to place 8 rooks on an 8×8 chessboard, then the number of all the possible placements is 64!/(64-8)! which is just 64-P-8 (k-permutation)
But, can't the same problem be approached as finding the number of possible ways we can give 64 places to 8 rooks, which I believe we should be able to solve using the multinomial coefficient ? If yes, what is the multinomial coefficient in this case?
Another problem which I believe can be solved by either k-permutations or the multinomial coefficient is finding the number of ways 52 cards can be dealt to 4 players.
Obviously using the multinomial coefficient this is just: 52! / 13!13!13!13!
Why doesn't (52-P-13)(52-P-13)(52-P-13)(52-P-13) give us the same answer?
Best Answer
Ok let's look at the card dealing example.
You are correct that the problem can be trivially solved by using multinomial coefficient. A good graphical representation of this can be found here: Multinomial Coefficient example
It is true that the correct answer is: $$\frac{52!}{13!13!13!13!}$$ So there are some issues with your permutation method......
The first problem with your permutation approach is that the permutations should be: $$P(52,13)*P(39,13)*P(26,13)*P(13,13)$$ instead of: $$P(52,13)*P(52,13)*P(52,13)*P(52,13)$$ because you can't deal the cards that have already been dealt to other people.
And the second problem is that the result from using multinomial coefficient does not care about the order at which the cards have been dealt (see link!). That means you will need combinations instead of permutations: $$C(52,13)*C(39,13)*C(26,13)*C(13,13)$$ expanding this gives you: $$\frac{52!}{39!13!}\frac{39!}{26!13!}\frac{26!}{13!13!}\frac{13!}{0!13!}$$ which is equivalent to the correct answer.