[SPLIT] how many 5-card hands have at most two red cards?

[Guest]

The game of euchre uses only the 9s, 10s, jacks, queens, kings, and aces from a standard deck of cards. How many five hand cards have AT LEAST two red cards?

Why won't the "exactly one" method work for this?

I did 12C2* 12C3= 14520

Thank you!

 

[galactus]

At least two is not the same as exactly one.

You could count up the number of ways of getting no reds and one red and

subtract that from the total number of possible hands.

Because the opposite of at least two reds is no reds and one red.

The total possible hands is \(\displaystyle \L\\C(24,5)=42,504\).

Now, how many ways to get no reds?.

\(\displaystyle \L\\\underbrace{C(12,5)}_{\text{5 blacks}}=?\)

How many ways to get one red?.

\(\displaystyle \L\\\underbrace{C(12,4)}_{\text{4 blacks}}*\underbrace{C(12,1)}_{\text{1 red}}=?\)

Therefore, the number of ways to get at least 2 reds is:

\(\displaystyle \L\\C(24,5)-(C(12,5)+C(12,4)*C(12,1))=?\)

 

[pka]

Re: [SPLIT] how many 5-card hands have at most two red cards

The game of euchre uses only the 9s, 10s, jacks, queens, kings, and aces from a standard deck of cards. How many five hand cards have AT LEAST two red cards?

At least two means two, three, four or five.
\(\displaystyle \L \sum\limits_{{\rm k = 2}}^{\rm 5} {12 \choose k}{12 \choose {5-k}}\)

 

[soroban]

Re: [SPLIT] how many 5-card hands have at most two red cards

Hello, Anna!

The game of Euchre uses the 9s, 10s, Js, Qs, Ks, and As of a standard deck of cards.
How many five-hand cards have AT LEAST two red cards?

Why won't the "exactly one" method work for this? . . . . explained below

I did 12C2* 12C3= 14520 . . . . no, that is exactly two reds

The opposite of "at least two red cards" is "less than two red cards".
. . That is, \(\displaystyle 0\) reds or \(\displaystyle 1\) red.

There are 24 cards: 12 red and 12 black.
There are \(\displaystyle \begin{pmatrix}24 \\5 \end{pmatrix}\:=\:42,504\) possible 5-card hands.

0 reds, 5 blacks: \(\displaystyle \:\begin{pmatrix}12 \\ 0\end{pmatrix}\cdot\begin{pmatrix}12 \\ 5\end{pmatrix} \:=\:792\)

1 red, 4 blacks: \(\displaystyle \;\;\begin{pmatrix}12 \\ 1\end{pmatrix}\cdot\begin{pmatrix}12 \\ 4\end{pmatrix} \:=\:5,940\)

. . Hence: \(\displaystyle \:n(\text{less than 2 red})\:=\:792\,+\,5,940\:=\:6,732\)

Therefore: \(\displaystyle \:n(\text{at least 2 red}) \:=\:42,504\,-\,6,732\:=\:\fbox{35,772}\)

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We can check our answer "the long way".

2 reds, 3 blacks: \(\displaystyle \:\begin{pmatrix}12 \\ 2\end{pmatrix}\cdot\begin{pmatrix}12 \\ 3\end{pmatrix}\:=\:14,520\)
3 reds, 2 blacks: \(\displaystyle \:\begin{pmatrix}12 \\ 3\end{pmatrix}\cdot\begin{pmatrix}12 \\2\end{pmatrix}\:=\:14,520\)
4 reds, 1 black: \(\displaystyle \;\:\begin{pmatrix}12\\4\end{pmatrix}\cdot\begin{pmatrix}12\\1\end{pmatrix} \:=\;\,5,940\)
5 reds, 0 black: \(\displaystyle \;\:\begin{pmatrix}{12\\5\end{pmatrix}\cdot\begin{pmatrix}12\\0\end{pmatrix}\:=\;\;\;792\)

Therefore: \(\displaystyle \:n(\text{at least 2 reds}) \:=\:14,520\,+\,14,520\,+\,5,940\,+\,792\:=\:\fbox{35,772}\)