Combinatorics: Ways to Choose Playing Cards

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To determine the number of ways to pick two different cards from a 52-card deck, where the first card is a spade and the second is not a queen, the approach involves two scenarios. If the first card is a non-queen spade, there are 12 options for the spade and 47 remaining cards to choose from, resulting in 12 x 47 combinations. If the first card is the queen of spades, there is 1 option for the spade and 48 remaining cards, yielding 1 x 48 combinations. The total number of combinations is then calculated by adding both scenarios together: (12 x 47) + (1 x 48). This method effectively resolves the initial confusion regarding the inclusion of the queen in the selection process.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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