How many five card hands can you have if?

[mHo2]

Homework Statement

if three of the cards have the same face value and the other two have different face values?

Homework Equations

nCr or (n!)/((n-r)!(r!))

The Attempt at a Solution

So far i have this for a solution

(13C1) * (4C3) * (12C1) * (4C1) * (11C1) * (4C1)

Where xCy is a combination, and 13 or 12 or 11 C 1 is the selection of the face value, and the 4c3 or 1 is the selection of the suit(clubs, hearts, diamonds, or spades).

But i have twice the value that it should be, I'm not sure what I'm doing wrong, if someone can explain this to me intuitively that would be fantastic!

I talked to my teacher and he mentioned something about arrangement, but the explanation didn't make much sense.

Thanks

 

[Hurkyl]

How many different 2-card hands are there in a deck of cards?

 

[mHo2]

How many different 2-card hands are there in a deck of cards?

I would say

52C2 so 1326

 

[Hurkyl]

To gain confidence in counting, we can always try to prove a one-to-one correspondence between the thing we are trying to count and the thing we actually count.

Just guessing from your product, what you are counting is the number of ways to do the following:

1. Pick a rank^*
2. Pick 3 different suits
3. Pick a rank different from step 1
4. Pick a suit
5. Pick a rank different from step 1 and step 3
6. Pick a suit

Try explicitly writing down how to convert these choices into a 5-card hand.

*: rank is the technical term for the "face value" of a card

 

[Hurkyl]

I would say

52C2 so 1326

Even though you are picking two cards, both different?

 

[mHo2]

Even though you are picking two cards, both different?

Oh, sorry! 52C1*51C1

To gain confidence in counting, we can always try to prove a one-to-one correspondence between the thing we are trying to count and the thing we actually count.

Just guessing from your product, what you are counting is the number of ways to do the following:

1. Pick a rank^*
2. Pick 3 different suits
3. Pick a rank different from step 1
4. Pick a suit
5. Pick a rank different from step 1 and step 2
6. Pick a suit

Try explicitly writing down how to convert these choices into a 5-card hand.

*: rank is the technical term for the "face value" of a card

Brilliant! That's exactly how I'm viewing it in my head. I'm not sure where my logic fails.

I suppose with the first two steps i would choose
13C1*4C3
This would pick a rank and then for the rank that was picked, choose three of the four suits for a combination
12C1*4C1
This would pick a different rank and then choose one of four suits (same for the 3rd step).

 

[Hurkyl]

Oh, sorry! 52C1*51C1

You were right the first time. I was trying to rephrase the question to make it sound more similar to the first problem. I had a hunch if I phrased it the other way you'd get the right answer -- I was hoping you'd catch both were asking the same question.

Brilliant! That's exactly how I'm viewing it in my head. I'm not sure where my logic fails.

Suppose my choices are:

1. Jack
2. Hearts, Spades, Clubs
3. Three
4. Spades
5. Seven
6. Diamonds

What is my hand?

When I say to explicitly write down the correspondence, I mean a direct algorithm one can use to answer questions like the above. An actual mathematical function (although not very numerical) that we can calculate with and prove things about.

The challenge is to prove:

• Every desired hand can be produced by this algorithm
• Two different lists of choices give two different hands

Or better yet, how to go in reverse. If my hand is

• Queen of Spades
• Three of Clubs
• Eight of Hearts
• Three of Diamonds
• Three of Spades

Then what six choices produce this hand?

Then check to see that the two algorithms are inverses of each other. Sometimes writing down the inverse and proving this is easier than the other option of proving onto and one-to-one.

 

[mHo2]

You were right the first time. I was trying to rephrase the question to make it sound more similar to the first problem. I had a hunch if I phrased it the other way you'd get the right answer -- I was hoping you'd catch both were asking the same question.

I'm having trouble differentiating between 52C1*51C1, and 52C2, why are they not the same thing?

Suppose my choices are:

1. Jack
2. Hearts, Spades, Clubs
3. Three
4. Spades
5. Seven
6. Diamonds

What is my hand?

Jack of Hearts
Jack of Spades
Jack of Clubs
Thee of Spades
Seven of Diamonds

When I say to explicitly write down the correspondence, I mean a direct algorithm one can use to answer questions like the above. An actual mathematical function (although not very numerical) that we can calculate with and prove things about.

The challenge is to prove:

• Every desired hand can be produced by this algorithm
• Two different lists of choices give two different hands

Or better yet, how to go in reverse. If my hand is

• Queen of Spades
• Three of Clubs
• Eight of Hearts
• Three of Diamonds
• Three of Spades

Then what six choices produce this hand?

Then check to see that the two algorithms are inverses of each other. Sometimes writing down the inverse and proving this is easier than the other option of proving onto and one-to-one.

I would say the 6 choices that produce this hand would be:
First rank is:
Three
The three accompanying suits are:
Diamonds, Spades, Clubs (not Hearts)
Next rank is:
Queen
Accompanying suit:
Spades
Last rank is:
Eight
Accompanying Suit:
Hearts

These seem like inverses!

 

[Hurkyl]

These seem like inverses!

Yes, they do seem like it. But you haven't actually stated an algorithm for either one yet...

(p.s. it's my bedtime. hopefully someone else will chime in if you need more help)

 

[mHo2]

Yes, they do seem like it. But you haven't actually stated an algorithm for either one yet...

(p.s. it's my bedtime. hopefully someone else will chime in if you need more help)

In that case, I'm not sure how to write an algorithm for it.
thanks for the help so far though!

 

[mHo2]

Okay so as an update my problem is:
I can't differentiate between (as an example) 52C1*51C1 and 52C2

what's the intuitive difference? selection group?

EDIT:
Apparently it's due to order not mattering. My problem here is, the solution to this question, which is

13C1*4C3*12C2*4C1*4C1

I'm not sure why I can't combine both of the 'rank picking' combinations if order doesn't matter, which it shouldnt! :S