Combinatorics: Ways to Choose Playing Cards

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Homework Help Overview

The discussion revolves around combinatorial methods for selecting playing cards from a standard 52-card deck, specifically focusing on the constraints of choosing one spade and ensuring the second card is not a queen.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore different scenarios based on whether the first card drawn is a queen or not, and how this affects the total number of valid selections for the second card.

Discussion Status

Participants are actively discussing the implications of their assumptions and calculations. Some have suggested a method for counting the different cases, while others are verifying the correctness of their reasoning and calculations.

Contextual Notes

There is an emphasis on ensuring that the second card drawn is not a queen, which complicates the counting process. Participants are also considering the impact of drawing the queen of spades as the first card.

Shoney45
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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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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.
 
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?
 
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.
 

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