# Thread: If you are dealing with a standard deck of cards-how many...data management

1. ## If you are dealing with a standard deck of cards-how many...data management

I don't know where this belongs but this seems close at least.

My partner and I have been going at this for over an hour. We are using nCr ish forumulas but this one doesn't seem to work...

If you are dealing from a standard deck of cards,
a)how many different 4-card hands could have at least one card from each suite
b)how many different 5-card hands could have at least one spade?
c) how many different 5-card hands could have at least 2 face cards(jacks, queens or kings)?

I am sorry if this is in the wrong board if if the title sucks. But thanks!

2. Hello, risteek!

You are dealing from a standard deck of cards,
a) How many 4-card hands have at least one card from each suit?
Strange wording . . . the "at least" is unnecessary.

We must have one card from each suit.

There are: 13 choices for the Club, 13 choices for the Heart,
. . 13 choices for the Spade, and 13 choices for the Diamond.

There are: . $13^4 \:=\:28,\!561$ possible hands.

b) How many 5-card hands have at least one spade?
There are 13 Spades and 39 Others.

There are: . $_{52}C_5 \:=\:2,\!598,\!960$ possible 5-card hands.

Then the five cards must be chosen from the 39 Other cards.
There are: . $_{39}C_5 \:=\:575,\!757$ ways.

There are: . $2,\!598,\!960 - 575,\!757 \:=\:2,\!023,\!203$ hands with at least one Spade.

c) How many 5-card hands have at least 2 face cards?
There are 12 Face cards and 40 Others.

The opposite of "at least 2 face cards" is "0 or 1 face card."

0 Face cards
The 5 cards are chosen from the 40 Others.
There are: . $_{40}C_5 \:=\:658,008$ ways to get 0 Face cards.

1 Face card
There are: . $_{12}C_1 = 12$ ways to get one Face card.
There are: . $_{40}C_4 = 91,\!390$ to get 4 Others.
. . There are: . $12\cdot91,\!390 \:=\:1,096,680$ ways to get one Face card.

Hence, there are: . $658,\!008 + 1,\!096,\!680 \:=\:1,\!754,\!688$ ways to get 0 or 1 Face card.

Therefore, there are: . $2,\!598,\!960 - 1,\!754,\!688 \:=\:844,\!272$ hands with at least 2 Face cards.

3. Thank you so much! You are amazing!!

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