Probability help: "Assuming you've got a deck of 40 cards..."

gooweegoo

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Jan 24, 2017
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Assuming you've got a deck of 40 cards (Ace, 2,3,4,5,6,7,jack,queen,king of 4 colors): How many hands of 10 cards can you make, where you need to have at least one jack and one king, that must be of the same color?

Here's how I tried to answer this question. The number of king/jack combinations you can make is 4. You then have 8 cards left, so you multiply 4 by the number of different possible hands of 8 cards possible, while taking in consideration there are only 38 cards left. You get 4((38!)/30!)

Then you must take in account the fact that the order of the cards doesn't matter. Thus you divide everything by 10!, the number of different permutations of the whole thing. So you have (4.38!)/((10!)(30!)) Which gives me a decimal number, which tells me I've done something wrong.

Anyone mind helping?
 

gooweegoo

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Jan 24, 2017
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Seems to me you can reword your problem:
40 cards, numbered 1 to 40, 4 colors:
1 to 10 : red
11 to 20 : white
21 to 30 : black
31 to 40 : blue
...or something similar; agree?
Yes.
 

ksdhart2

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Mar 25, 2016
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The basic idea's there, but your execution and reasoning is a bit off. Starting from the very beginning, there's actually more than 4 king/jack combinations. You're correct that order doesn't matter here, so let's just say the jack always goes in the first slot. If it's the Jack of Spades, there's four possible Kings that can go in the second slot. But you say the Jack and King need to be of the same color (did you possibly mean the same suit, or would Clubs and Spades be sufficient?). So, how many of those Kings match the Jack's color? Now let's say instead the first slot has the Jack of Clubs. Same question - how many of the Kings match the Jack's color? Repeating for all four possible suits of Jacks, how many of the 16 total Jack/King combinations are there that have the same color?

Now, you need to fill out the remaining eight slots by picking cards from the deck. You started with 40 cards, so there's now 38 remaining. Order doesn't matter, nor does which specific cards get picked, any eight will do. So, how many ways can you pick 8 cards out of a deck of 38? Is there perhaps a formulahttps://www.mathsisfun.com/combinatorics/combinations-permutations.html you know for this?
 
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