Question

Five cards are dealt from a standard deck of 52 cards.
a) How many different hands can be dealt?
b) How many hands will contain exactly 4 hearts (and 1 other card)?
c) How many hands will contain no face cards?
d) How many hands will contain only spades or only diamonds?
e) How many hands will contain only red cards?
f) How many hands will contain no more than 2 hearts?

Answer 1

Are you familiar with binomial coefficients?

\[\left(\begin{matrix}N \\ k\end{matrix}\right) = \frac{N!}{(N-k)!k!}\]

Answer 2

Uh no, sorry. I'm have grade 12 data management and I don't think I've learned that yet. I'm currently doing permutations and combinations

Answer 3

It's a version of those :)

The binomial coefficient tells us how many ways there are to pick k things from N things, without replacement, regardless of order.

So:

\[\left(\begin{matrix}52 \\ 5\end{matrix}\right) = \frac{52!}{(52-5)!5!} = \frac{52*51*50*...*3*2*1}{(47*46*...*3*2*1)*5*4*3*2*1} = \frac{52*51*50*49*48}{5*4*3*2*1}\]

Tells us the number of ways to choose 5 cards from 52 (This is the total number of hands possible).

This works rather simply. If we were choosing 52 cards from a deck of 52, and order mattered, we'd have 52! total combinations. We only want 5, so we remove 52-5=47 choices, leaving us at:
\[\frac{52!}{(52-5)!} = \frac{52!}{47!}\]
We then remove the stipulation or order mattering by dividing be the number of ways a given 5 cards could be arranged. That is, we divide by 5!, since we are drawing 5 cards.

What we're left with is the binomial coefficient.

If you're ok with using this, this problem is pretty easy. If you'd rather not, I'll have to think about how to do it a bit :)

Answer 4

hmm. that seems confusing lol

Answer 5

And just for clarification, the end of that long line got chopped off:
\[\left(\begin{matrix}52 \\ 5\end{matrix}\right) = \frac{52*51*50*49*48}{5*4*3*2*1}\]

Answer 6

Hmmm, without that, it's a long walk. Let's try, though.

First, how many different hands of 5 cards can be made?
For the first card, we can draw any one of the 52 cards. For the second card, we only have 51 to choose from. And it keeps decreasing for each card drawn. How many combinations could we make?

Answer 7

311,875,200?

Answer 8

i did 52x51x50x49x48

Answer 9

Right, but that's if order matters. 52*51*50*49*48.

We don't care if we get the Ace of Spades first, second, or last. We just care that we get it. Right? So we have to remove all the different combinations that give the same hand.

Do you know how to do this?

Answer 10

52c5?

Answer 11

52 choose 5?!?! That's the binomial coefficient :D

Answer 12

And yes, that's how you'd do it

Answer 13

so, there are 2,598,960 different hands that can be dealt?

Answer 14

Yup!

Are you ok with using the 52c5 notation? If so, this just got a lot easier.

\[52c5 = \left(\begin{matrix}52 \\ 5\end{matrix}\right)\]
Which is what I was wanting to use :)

Answer 15

oh.

Answer 16

yeah, i guess i am okay with it now

Answer 17

before we move on, can i ask you about something else?

Answer 18

Excellent. So, you're done with a.

Now, we want to insist on 4 cards being a heart, and 1 card that isn't a heart.

Answer 19

Sure, whats up?

Answer 20

(11+8 choose 2). do i add the 11 after or is it 19c2?

Answer 21

this is unrelated to this question btw

Answer 22

i just wanted clarification on this before moving on

Answer 23

I'll do my best to answer it :)

Answer 24

do u want the full question? i asked it last night.

Answer 25

not sure if you can see it but here http://openstudy.com/users/dudehao#/updates/5362de59e4b02f72432c1ee6

Answer 26

The chemists and technicians problem? Alrighty. What would you like clarified?

Answer 27

the 11+8choose2. is it 19c2 or 8c2 then i add 11 to it

Answer 28

With the notation that was being used, it would be 19c2

I'm checking to make sure I get the same answer for d, though.

Answer 29

oh okay, thanks

Answer 30

let me know once youre finished

Answer 31

Yeah. And he did it in a much simpler way than what I started doing initially :)

Answer 32

You do it by first insisting that you get at least 1 technician: 12c1 = 12
Then insist that you get at least 1 chemist: 9c1 = 9
The other two, we don't really care about. We just need 2 of them. There were 21 total people. We've chosen 2, so we have 19 left to choose from: 19c2

(12c1)(9c1)(19c2) = 12*9*(19c2)

Answer 33

oh wow. thanks for clearing that up. now let's move on to the deck question

Answer 34

Well, we'll approach this in the same way.

First, we want to insist that we choose 4 hearts. There are 4 suits in a deck. So to specify that we want a heart, we do: (4c1)

We then want 4 of these hearts. There are 13 hearts (13 cards of each suit) in a deck. Since we've already chosen our suit, we just pick 4 of them: (13c4)

Then we just have to specify that the other card must come from another suit. How would we do this?

Answer 35

uhm, 52c1?

Answer 36

or 13c1?

Answer 37

Close. We have to insist that it is NOT a heart, as well. 52c1 gives us a single random card from the deck. Could be a heart. Could not be. 13c1 gives us a single card from a suit, but it doesn't limit what that suit could be.

Answer 38

39c1?

Answer 39

That's one way to do it, sure :D

Another way: Insist that the card is one of the other three suits: 3c1
Draw one from that suit: 13c1

Then we multiply it all together.

(4c1)(13c4)(39c1) = ?

Answer 40

I got 111,540

Answer 41

Yep :)

Answer 42

Yes, thank you

Answer 43

Now, how would you do the next one?

Answer 44

how would i do c? what is a face card exactly?

Answer 45

king queen and jack?

Answer 46

Right. King, Queen, and Jack are face cards. You can use a choose method. There are also other methods. Try it out, let's see what you get :)

Answer 47

for c, do I do 38c5?

Answer 48

Why 38?

Answer 49

because there are 14 facecards, i subtracted that from 52

Answer 50

Jack Queen and King of each suit. 4 suits.
3*4 = 12 face cards

Answer 51

so jokers dont count?

Answer 52

Not in a 52 card deck. There are no jokers, unless they change the deck somehow.

The cards are: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K
of each suit. 13 cards in each suit.
13*4 = 52

Answer 53

ok i see, so it would be 36c5?

Answer 54

52-12=40

Answer 55

whoops true ahah

Answer 56

So, 40c5 ways to make hands without face cards.

Answer 57

yup, and i get 658,008

Answer 58

Yep.

What about the next part?

Answer 59

26c5?

Answer 60

only spades OR only diamonds. Not a combination of spades and diamonds.

Answer 61

13c5?

Answer 62

First, how would we decide how many hands could be made with only spades?

Answer 63

misread the question whoops

Answer 64

Ok, 13c5 will gives us a hand all from a single suit. So if we wanted this OR the same thing from another suit, we'd just add the probability to do it with the other suit to this.

How many ways are there to make hands with only diamonds?

Answer 65

it would also be 13c5?

Answer 66

Yep. So (13c5)+(13c5) = 2*(13c5) = ?

Answer 67

2574 is my answer

Answer 68

Alright. And the next part?

Answer 69

26c5

Answer 70

Good :)

Answer 71

thanks, you made this so much easier. yesterday i had no clue what I was doing lol

Answer 72

Just a quick note: I made a mistake on part b

The (4c1) shouldn't be there. We did a slightly different problem.

So for b, you'd have: (13c4)(39c1)

Answer 73

alright thanks and for the last part do i do 13c2?

Answer 74

For the last part, we want NO MORE THAN 2 hearts.

So, we can have no hearts, 1 heart, or 2 hearts.

Find the number of ways to make a hand with no hearts. And the number of ways to make a hand with only 1 heart. And finally, with only 2 hearts. Then add them all up.

Answer 75

39c5 + 13c1 + 13c2?

Answer 76

Close.

Remember, we have to draw 5 cards each time. 39c5 is fine, as you chose 5.

13c1, you only drew one card. We need to draw 4 more that are NOT hearts.
13c2, we need to draw 3 more.

Answer 77

so 13c5 x2?

Answer 78

oh okay

Answer 79

So, if I want only a single heart:
I draw that heart (13c1)
Then I draw 4 more cards that are NOT hearts. number of cards that aren't hearts: 52-13=39
So draw 4 that are not hearts: (39c4)

So the number of ways to make a hand with only one heart is:
(13c1)(39c4)

Answer 80

so then the number of ways to make two hearts would be 13c2 x 39c3?

Answer 81

Yep :)

Answer 82

so i get 795,093 in total

Answer 83

I gotta head out now, but thank you for everything. I'll let you know if I need help with any more questions, thanks again!

Answer 84

That's not what I get