In how many ways can 12 prices be awarded evenly among four people?
I am not sure how to start this question.
Also, one more:
From a deck of 52 cards, how many different four-card hands could be dealt which include one from each suit?
Again, I do not know where to start. Additionally, if anyone can link me to a source with more of these combination questions (deck of card questions), that would be great.
Warning My responses to combinations questions are notoriously inaccurate
How many ways can 12 prizes be awarded amongst 4 people.
You know that each person gets 3 prizes.
There are 12! ways of distributing the prizes, and each person gets 3 of them. But the order of the 3 prizes receieved doesn't matter. there are 3! ways of ordering the set of 3 prizes receieved
From a deck of 52, how many 4 card hands include 1 from each suit
There are 4 suits in a standard deck of cards. So you must choose 1 of the 13 spades, then 1 of the 13 clubs,....etc
There are 13*13*13*13 ways of doing that. However the order does not matter so the final answer is
13*13*13*13 / 4!
(there are 4! ways of ordering the 4 suits)
Thanks for the help, however, these are inaccurate. Thanks, though.
Originally Posted by SpringFan25
12 prizes can be divided evenly among 4 people in
Originally Posted by NeedsHelpp
12! / (3!)^4
ways. One way to look at this is as a "multinomial coefficient". Another way is that the prizes can be ordered in 12! ways with person #1 getting 1-3, #2 getting 4-6, etc., but the ordering of gifts for each person doesn't matter.
For the card problem, there are 13 ways to pick a club, 13 ways to pick a spade, etc., so there are 13^4 possible hands. (We don't need to divide by 4! because we specified the suites.)
i think ill stop answering these ones :D