Combinatorics: Ways to Choose Playing Cards

In summary: So the total number of ways to pick 2 different cards from a standard 52-card deck such that the first card is a spade and the second card is not a queen is 588. In summary, there are 588 ways to pick 2 different cards from a standard 52-card deck such that the first card is a spade and the second card is not a queen. This is calculated by adding the number of ways to pick a non-queen spade and a non-queen card (12 x 47) to the number of ways to pick the queen of spades and a non-queen card (1 x 48).
  • #1
Shoney45
68
0

Homework Statement


How many ways are there to pick 2 different cards from a standard 52-card deck such that the first card is a spade and the second card is not a queen.


Homework Equations



P(n,k) C(n,k)

The Attempt at a Solution



Since the player must have one spade, there are 13 ways to deal a spade. I now have 51 cards to deal out.

I am mixed up on what to do with the second card though. I can't find a way to deal with the case where the first spade that I deal out happens to be a queen. If it was a queen, then I still need to take out three other queens from the deck, leaving me with 48 cards. If the first card wasn't a queen, then I still need to take out four cards from the deck, leaving me with 47 cards. 13 x 48 is a different result than 13 x 47. But I know that I am only supposed to have one answer. So I really don't know how to proceed on this one.
 
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  • #2
Shoney45 said:

Homework Statement


How many ways are there to pick 2 different cards from a standard 52-card deck such that the first card is a spade and the second card is not a queen.


Homework Equations



P(n,k) C(n,k)

The Attempt at a Solution



Since the player must have one spade, there are 13 ways to deal a spade. I now have 51 cards to deal out.

I am mixed up on what to do with the second card though. I can't find a way to deal with the case where the first spade that I deal out happens to be a queen. If it was a queen, then I still need to take out three other queens from the deck, leaving me with 48 cards. If the first card wasn't a queen, then I still need to take out four cards from the deck, leaving me with 47 cards. 13 x 48 is a different result than 13 x 47. But I know that I am only supposed to have one answer. So I really don't know how to proceed on this one.

Note that the first card is either a Queen of Spades or it isn't. Thus, to count the total number of ways to do this, you can count how many ways there are to pick cards such that the first card is a non-queen spade and the second is a non-queen, then count the number of ways there are to pick the queen of spade and then a non-queen card. Then add those two number together.
 
  • #3
Robert1986 said:
Note that the first card is either a Queen of Spades or it isn't. Thus, to count the total number of ways to do this, you can count how many ways there are to pick cards such that the first card is a non-queen spade and the second is a non-queen, then count the number of ways there are to pick the queen of spade and then a non-queen card. Then add those two number together.

Okay, I think I got it then. So if I give the player a spade that does not happen to be a queen, and then I take out the four queens, I am left with (12 x 47). If he happens to receive the queen of spades first, and I take out the remaining three queens, that equals (1 x 48). So the answer is (12 x 47) + (1 x 48). Does that look about right?
 
  • #4
Shoney45 said:
Okay, I think I got it then. So if I give the player a spade that does not happen to be a queen, and then I take out the four queens, I am left with (12 x 47). If he happens to receive the queen of spades first, and I take out the remaining three queens, that equals (1 x 48). So the answer is (12 x 47) + (1 x 48). Does that look about right?

Yep, looks right to me.
 

1. How many ways can a standard deck of 52 playing cards be arranged?

There are 52! (52 factorial) ways to arrange a standard deck of 52 playing cards. This means that there are 52 options for the first card, 51 options for the second card, and so on, until there is only 1 option for the last card. The total number of arrangements is equal to 52 x 51 x 50 x ... x 2 x 1 = 52! = 8.0658 x 10^67.

2. How many possible hands can be dealt in a game of poker?

There are 2,598,960 possible hands that can be dealt in a game of poker. This is calculated by using the combination formula, nCr = n! / (r! * (n-r)!), where n is the number of items to choose from and r is the number of items to be chosen. In poker, there are 52 cards in a deck and a hand consists of 5 cards, so the formula becomes 52C5 = 52! / (5! * (52-5)!) = 2,598,960.

3. How many ways can a hand of 5 cards be dealt from a deck of 52 cards?

There are 2,598,960 ways a hand of 5 cards can be dealt from a deck of 52 cards. This is calculated using the same combination formula as in question 2, where n = 52 and r = 5.

4. How many ways can a hand of 5 cards be dealt from a deck of 52 cards if order does not matter?

If order does not matter, then the number of ways to deal a hand of 5 cards from a deck of 52 cards is equal to the number of combinations, which is calculated using the formula nCr = n! / (r! * (n-r)!). So the number of combinations of 52 cards taken 5 at a time is 52C5 = 52! / (5! * (52-5)!) = 2,598,960.

5. How many ways can a hand of 5 cards be dealt from a deck of 52 cards if order matters?

If order matters, then the number of ways to deal a hand of 5 cards from a deck of 52 cards is equal to the number of permutations, which is calculated using the formula nPr = n! / (n-r)!. So the number of permutations of 52 cards taken 5 at a time is 52P5 = 52! / (52-5)! = 52! / 47! = 311,875,200.

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