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January 24th, 2017, 10:31 AM   #1
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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?
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January 24th, 2017, 01:32 PM   #2
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There is at least one error in your analysis. Hands containing more than one combination are counted more than once.
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January 24th, 2017, 05:07 PM   #3
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you have to break it up into how many jack/king pairs appear

$n = \begin{pmatrix}4 \\ 1\end{pmatrix}\begin{pmatrix}32 \\ 8\end{pmatrix} +
\begin{pmatrix}4 \\ 2\end{pmatrix}\begin{pmatrix}32 \\ 6\end{pmatrix} +
\begin{pmatrix}4 \\ 3\end{pmatrix}\begin{pmatrix}32 \\ 4\end{pmatrix} +
\begin{pmatrix}4 \\ 4\end{pmatrix}\begin{pmatrix}32 \\ 2\end{pmatrix} =47654688$
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January 25th, 2017, 05:21 PM   #4
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Quote:
Originally Posted by romsek View Post
you have to break it up into how many jack/king pairs appear

$n = \begin{pmatrix}4 \\ 1\end{pmatrix}\begin{pmatrix}32 \\ 8\end{pmatrix} +
\begin{pmatrix}4 \\ 2\end{pmatrix}\begin{pmatrix}32 \\ 6\end{pmatrix} +
\begin{pmatrix}4 \\ 3\end{pmatrix}\begin{pmatrix}32 \\ 4\end{pmatrix} +
\begin{pmatrix}4 \\ 4\end{pmatrix}\begin{pmatrix}32 \\ 2\end{pmatrix} =47654688$
You are omitting hands where there is a j-k pair plus unpaired j's or/and k's.
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January 30th, 2017, 08:30 AM   #5
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Let n be a natural number and a multiple of 4. There is a deck D of n cards where each card has a unique rank and unique color (suit). For each rank, there are exactly 4 cards of the same rank of different colors. Given two different ranks, say kings and jacks, how many hands of 10 cards from D have at least one king and one jack of the same color?

For counting problems like this one, there is a pretty high probability that I'll make an error. (I don't think I'm the only mathematician that routinely makes counting errors.) It is for this reason that I rephrased your problem to a general n. This way I could have the computer generate all possible hands from D and then count the number of desired hands. I can then "verify" any supposed formula. A priori, I was afraid a 40 card deck would take too long for the generation; it turned out my PC could generate all possible hands from a 40 card deck in about 30 seconds. The answer T for a 40 card deck is 184,112,224; for a 52 card deck it's T=2,074,537,322 (52 card deck not verified by examining all hands).

Let C(m,r) be the number of combinations of m things taken r at a time (the number of r sized subsets from a set with m elements). Here are two solutions; the first of which is from the previous two posts.

1. Let $H_i$ be the set of desired hands with exactly $i$ king jack pairs of the same color and $h_i=|H_i|$. Clearly $T=h_1+h_2+h_3+h_4$. Now
$$h_1=C(4,1)\sum_{j=0}^32^jC(3,j)\,C(n-8,8-j)$$
Let K be the set of kings and jacks, |K|=8. The first factor C(4,1) is the number of king jack pairs of the same color; the jth term of the sum is the number of ways of choosing j cards from the remaining 6 cards of K with no king jack pair of the same color and then 8-j cards from $D\setminus K$.
Similarly:
$$h_2=C(4,2)\sum_{ j=0}^22^jC(2,j)\,C(n-8,6-j)$$
$$h_3=C(4,3)\sum_{ j=0}^12^jC(1,j)\,C(n-8,4-j)$$
$$h_4=C(4,4)C(n-8,2)$$

2. Again K is the set of 8 kings and jacks. Pretty obviously (pigeon hole principle if nothing else) a hand with at least 5 cards from K has at least one king jack pair of the same color. Let $a_i$ be the number of subsets of K with $i$ elements that contain at least one king jack pair of the same color. Then

$$T=a_2C(n-8, +a_3C(n-8,7)+a_4C(n-8,6)+\sum_{ j=5}^8C(8,j)C(n-8,8-j)$$
Clearly $a_2=4$ and $a_3=4\cdot 6$. Now for $a_4$, the number of 4 element subsets of K that do not contain a king jack pair of the same color is 16. So $a_4=C(8,4)-16$.

Last edited by johng40; January 30th, 2017 at 08:35 AM.
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