How many ways are there to deal the playing cards in a 52-card deck to four different players so that each player gets an equal number of cards, one player gets all the spades, and each player gets a king?
My Answer: C(4, 1) * 3! * P(36; 12,12,12) My Reasoning: C(4, 1) ways of dealing the 13 spades, 3! ways of distributing the kings among the remaining three people, P(36;12,12,12) ways of distributing 36 distinct objects into 3 separate boxes with 12 in each box.
Can someone check my answer, and if its wrong, explain why?
The really irritating thing about these questions is it's not at all clear what is a "way" of dealing cards. You seem to be assuming that the order in which the the player gets the spades doesn't matter, nor does it matter when another player gets a king which of his thirteen cards is the king. But then if those don't matter does it matter what player gets the card? And does it matter if which of the other players get which kings! Frankly, this is way too ambiguous. Each interpretation has a specific answer but it's not at all clear which interpretation matters.
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Your way is correct if we assume it does matter which player is signalled for all spades but it doesn't matter what order the spades come in; the it does matter which player gets which king but not which position the king comes in, and ....I'm not sure what $P(36:12, 12,12)$ means but it seems like you are claiming the what the other $12$ cards and order each player gets and order does matter? Or doesn't?
I figure order either matters for everyone or it doesn't matter for anyone.
So if it matters which player gets all spades, there are $C(4,1) ={4 \choose 1}=4$ ways for that. But order doesn't matter. If it matters which of the other $3$ get which king there are $3!$ ways to choice which player gets which king. Of the remaining cards, We arbitrarily pick a player one and there are $C(36,12)={36\choose 12}$ for her cards and arbitrairly pick player two and there are $C(24,12)$ and $C(12,12)$ for the third.
When you write $P(N; a,b,c,...)$ does that mean ${N\choose a}{N-a\choose b}{N-a-b \choose c}....$ which will indeed equal $\frac{N!}{a!b!c!.....}$? So I suppose you do mean that.