Combinations of cards in a hand of euchre

The game of euchre uses only 9s, 10s, jacks, queens, kinds, and aces from a standard deck of cards. How many five-card hands have:
(a) all red cards?
(b) at least two red cards?
(c) at most two red cards?

Re: Combinations

kimmy_koo51 said:

The game of euchre uses only 9s, 10s, jacks, queens, kinds, and aces from a standard deck of cards. How many five-card hands have:
(a) all red cards? (b) at least two red cards? (c) at most two red cards?

Click to expand...

The deck of ‘euchre’ has 24 cards.
a) Half of those are red \(\displaystyle {{12} \choose 5}.\)
b) \(\displaystyle \sum\limits_{k = 2}^5 {{12} \choose k}{{12} \choose {5-k}}\)

There are 6 denominations, therefore, 12 red cards. There are 2 red cards per denomination, hearts and diamonds.

So, if the hand is all red, we must select 5 cards from the 12 red. C(12,5)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

C(24,5)-(C(12,0)C(12,5)+C(12,1)C(12,4))

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

At most 2 red cards:

C(12,0)C(12,5)+C(12,1)C(12,4)+C(12,2)C(12,3)


Your task is to interpret. See how they work?.

A true Canadian does not play euchre, but plays "500" :shock:

I spent many winter nights playing 500 with my 3 brothers:

"Canadian Five Hundred
500 is played, under the name Cinq Cents, in French Canada, especially Montreal. It is played with a 46-card pack, made by throwing out the twos and threes from a 52-card pack and adding two distinguishable jokers, which are the highest trumps. There are four players in fixed partnerships, each of whom receive 10 cards; the high bidder takes the kitty of six cards and discards six."

Of course we developed forms of cheating, like put a card down sideways if
we wanted our partner to come back in same suit :idea:

Hello, kimmy_koo51!

The game of euchre uses only 9s, 10s, jacks, queens, kinds, and aces
from a standard deck of cards.
How many five-card hands have:

(a) all red cards?
(b) at least two red cards?
(c) at most two red cards?

Click to expand...

There are: \(\displaystyle \:{24\choose5} \:=\:\frac{24!}{5!19!} \:=\:42,504\) possible hands.


(a) pka is absolutely right.
There are 12 red cards and we must select 5 of them.
. . There are: \(\displaystyle \:{12\choose5} \:=\:\frac{12!}{5!7!} \:=\:792\) possible all-red hands.


(b) "At least two red cards" (two or more red cards)
The opposite of "2 or more red cards" is "less than two red cards".
. . This means: "0 red cards" or "1 red card".

"No red" means "all black"; we must choose 5 black cards.
. . \(\displaystyle n(0\text{ red})\:=\:n(\text{5 black}) \:=\:{12\choose5} \:=\:792\) ways.

"1 red (and 4 black)"
. . \(\displaystyle n(\text{1 red, 4 black}) \:=\:{12\choose1}{12\choose4}\:=\:5940\) ways.

Hence, there are: \(\displaystyle \,792\,+\,5940\:=\:6732\) ways to get less than two reds.

Therefore: \(\displaystyle n(\text{2 or more red})\:=\:42,504\,-\,6,732\:=\:35,772\) ways. **


(c) "At most two red cards"
This means: "0 red", "1 red", or "2 red".

From part (b) we have: \(\displaystyle n(\text{0 red}) \:=\:792,\;n(\text{1 red}) \:=\:5,940\)

We need "2 red (and 3 black)".
. . \(\displaystyle n(\text{2 red, 3 black}) \:=\:{12\choose 2}{12\choose 3} \:=\:\frac{12!}{2!10!}\cdot\frac{12!}{3!9!} \:=\:14,520\)


Therefore: \(\displaystyle \,n(\text{at most 2 red}) \:=\:792\,+\,5,940\,+\,14,520 \:=\:21,252\) ways.

**
Edit: corrected a silly subtraction error . . .

soroban said:

(a) pka is absolutely right.

Click to expand...

Well thank you. It is a great relief to me to have your approval.
But please tell me, is the rest of the post showing the student how to do the question or is a complete working out of the question? If it is the latter, then I am sure you will be voted the best helper here. After all, who wants to learn if the job is just handed to me?