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Number of hands with $13$ cards that contain at least one picture card; $4$ cards of the same rank; one card of each rank

0

$begingroup$

Find the number of hands of $13$ cards that contain:

(a) at least $1$ picture card (where a picture card has rank J, Q, K or A)

(b) $4$ cards of the same rank

(c) one card of each rank

For (a), I came up with answer as $frac1652 times frac1551 times...timesfrac540 $. Is there a quicker way to calculate this?

I also have in mind that I get to choose $13$ cards.. so there might be a different way to solve these kinds of questions?
Any help on how to tackle (b) and (c) will be appreciated!

I am a little confused as there are $13$ cards to draw but need to find the number of hands for one card of each rank.

combinatorics

share|cite|improve this question

edited Mar 20 at 10:34

N. F. Taussig

45k103358

asked Mar 19 at 12:05

MathsGoogleMathsGoogle

465

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For (a), you are calculating the probability that you have a hand of only picture cards. You were supposed to calculate the number of hands with at least one picture card.
  $endgroup$
  – Arthur
  Mar 19 at 12:15
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Find the number of hands of $13$ cards that contain:

(a) at least $1$ picture card (where a picture card has rank J, Q, K or A)

(b) $4$ cards of the same rank

(c) one card of each rank

For (a), I came up with answer as $frac1652 times frac1551 times...timesfrac540 $. Is there a quicker way to calculate this?

I also have in mind that I get to choose $13$ cards.. so there might be a different way to solve these kinds of questions?
Any help on how to tackle (b) and (c) will be appreciated!

I am a little confused as there are $13$ cards to draw but need to find the number of hands for one card of each rank.

combinatorics

share|cite|improve this question

edited Mar 20 at 10:34

N. F. Taussig

45k103358

asked Mar 19 at 12:05

MathsGoogleMathsGoogle

465

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For (a), you are calculating the probability that you have a hand of only picture cards. You were supposed to calculate the number of hands with at least one picture card.
  $endgroup$
  – Arthur
  Mar 19 at 12:15
  
  

  
  
  
  

add a comment | 

$begingroup$

Find the number of hands of $13$ cards that contain:

(a) at least $1$ picture card (where a picture card has rank J, Q, K or A)

(b) $4$ cards of the same rank

(c) one card of each rank

For (a), I came up with answer as $frac1652 times frac1551 times...timesfrac540 $. Is there a quicker way to calculate this?

I also have in mind that I get to choose $13$ cards.. so there might be a different way to solve these kinds of questions?
Any help on how to tackle (b) and (c) will be appreciated!

I am a little confused as there are $13$ cards to draw but need to find the number of hands for one card of each rank.

combinatorics

share|cite|improve this question

edited Mar 20 at 10:34

N. F. Taussig

45k103358

asked Mar 19 at 12:05

MathsGoogleMathsGoogle

465

$endgroup$

share|cite|improve this question

edited Mar 20 at 10:34

N. F. Taussig

45k103358

asked Mar 19 at 12:05

MathsGoogleMathsGoogle

465

share|cite|improve this question

edited Mar 20 at 10:34

N. F. Taussig

45k103358

asked Mar 19 at 12:05

MathsGoogleMathsGoogle

465

edited Mar 20 at 10:34

N. F. Taussig

45k103358

edited Mar 20 at 10:34

N. F. Taussig

45k103358

asked Mar 19 at 12:05

MathsGoogleMathsGoogle

465

asked Mar 19 at 12:05

MathsGoogleMathsGoogle

465

2 Answers
2

active

oldest

votes

1

$begingroup$

For $(a)$, the number of such hands can be calculated as follows:$$X=textnumber of all possible hands\-textnumber of all possible hands with no picture card\=binom5213-binom52-4times 413$$

For $(b)$, we choose a 4-tuple of same-rank cards in $13$ different ways, leaving the other cards to be chosen independently; totally$$Y=13timesbinom52-413-4$$

Edit

In case you want exactly 1 4-tuple, the answer is $$Y=13timesbinom489-binom132timesbinom445$$

For $(c)$, it is simply $$Z=4^13$$(why?)

share|cite|improve this answer

edited Mar 19 at 12:26

answered Mar 19 at 12:18

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For b) I think you are over-counting hands with more than one set of four.
  $endgroup$
  – Especially Lime
  Mar 19 at 12:22
  

  
  
  
  


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  $begingroup$
  So, do you think the question has restricted to only $1$? Anyway, thank you for giving me that feedback. I added the special case too...
  $endgroup$
  – Mostafa Ayaz
  Mar 19 at 12:23
  
  

  
  
  
  


• 
  
  
  

  
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  @MostafaAyaz: No, that is not the problem. You are counting hands with e.g. four Threes and four Sevens twice. So your final answer is too large. (I think you should remove your edit $-$ it is not relevant.)
  $endgroup$
  – TonyK
  Mar 19 at 12:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. I don't understand. Why twice? First I count the number of hands containing at least one 4-tuple. This is possible in 13 different ways. After removing that 4-tuple, we are left with $48$ more cards from which we are supposed to choose the other $9$ cards of ours. This is possible in $binom489$. In case at least 2 four-tuples are to be drawn out, nothing is counted twice since I used $nCr$.
  $endgroup$
  – Mostafa Ayaz
  Mar 19 at 12:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MostafaAyaz the point is that one of the options your method counts is to remove all four threes, then choose nine more cards which are A,K,Q,J,10 of spades and all the sevens. Another option you've counted is to remove all four sevens, then choose nine more cards consisting of A,K,Q,J,10 of spades and four threes. But these two options are the same hand.
  $endgroup$
  – Especially Lime
  Mar 19 at 13:28
  

  
  
  
  

add a comment | 

0

$begingroup$

For a) what you have is wrong on two counts: it is a probability, not a number of hands, and it is the probability of all cards being pictures.

The first point is easily fixed: just multiply by the total number of hands which is $binom5213$. The easiest way to correct the second point is to notice that this method can be used to calculate the probability of all cards being non-pictures: subtract that from 1 to get the probability of at least one picture.

share|cite|improve this answer

answered Mar 19 at 12:17

Especially LimeEspecially Lime

22.7k23059

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that for (a), he's calculating including ordering, meaning it's not $binom5213$.
  $endgroup$
  – Arthur
  Mar 19 at 12:18
  

  
  
  
  


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  $begingroup$
  @Arthur The question just asks for the number of hands of this type, and that normally means different orderings count as the same hand.
  $endgroup$
  – Especially Lime
  Mar 19 at 12:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I agree that that's probably what the problem means. But that's not the OP's approach, as can be seen from the $frac1652times frac1551timescdots$ calculation, meaning your answer ishas not adressed all the issues with (a)
  $endgroup$
  – Arthur
  Mar 19 at 12:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur if you're calculating probabilities, it doesn't matter whether you take account of ordering or not, you get the same answer. So I don't think this is really an issue.
  $endgroup$
  – Especially Lime
  Mar 19 at 13:24
  

  
  
  
  

add a comment | 

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1

$begingroup$

For $(a)$, the number of such hands can be calculated as follows:$$X=textnumber of all possible hands\-textnumber of all possible hands with no picture card\=binom5213-binom52-4times 413$$

For $(b)$, we choose a 4-tuple of same-rank cards in $13$ different ways, leaving the other cards to be chosen independently; totally$$Y=13timesbinom52-413-4$$

Edit

In case you want exactly 1 4-tuple, the answer is $$Y=13timesbinom489-binom132timesbinom445$$

For $(c)$, it is simply $$Z=4^13$$(why?)

share|cite|improve this answer

edited Mar 19 at 12:26

answered Mar 19 at 12:18

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For b) I think you are over-counting hands with more than one set of four.
  $endgroup$
  – Especially Lime
  Mar 19 at 12:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, do you think the question has restricted to only $1$? Anyway, thank you for giving me that feedback. I added the special case too...
  $endgroup$
  – Mostafa Ayaz
  Mar 19 at 12:23
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MostafaAyaz: No, that is not the problem. You are counting hands with e.g. four Threes and four Sevens twice. So your final answer is too large. (I think you should remove your edit $-$ it is not relevant.)
  $endgroup$
  – TonyK
  Mar 19 at 12:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. I don't understand. Why twice? First I count the number of hands containing at least one 4-tuple. This is possible in 13 different ways. After removing that 4-tuple, we are left with $48$ more cards from which we are supposed to choose the other $9$ cards of ours. This is possible in $binom489$. In case at least 2 four-tuples are to be drawn out, nothing is counted twice since I used $nCr$.
  $endgroup$
  – Mostafa Ayaz
  Mar 19 at 12:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MostafaAyaz the point is that one of the options your method counts is to remove all four threes, then choose nine more cards which are A,K,Q,J,10 of spades and all the sevens. Another option you've counted is to remove all four sevens, then choose nine more cards consisting of A,K,Q,J,10 of spades and four threes. But these two options are the same hand.
  $endgroup$
  – Especially Lime
  Mar 19 at 13:28
  

  
  
  
  

add a comment | 

1

$begingroup$

For $(a)$, the number of such hands can be calculated as follows:$$X=textnumber of all possible hands\-textnumber of all possible hands with no picture card\=binom5213-binom52-4times 413$$

For $(b)$, we choose a 4-tuple of same-rank cards in $13$ different ways, leaving the other cards to be chosen independently; totally$$Y=13timesbinom52-413-4$$

Edit

In case you want exactly 1 4-tuple, the answer is $$Y=13timesbinom489-binom132timesbinom445$$

For $(c)$, it is simply $$Z=4^13$$(why?)

share|cite|improve this answer

edited Mar 19 at 12:26

answered Mar 19 at 12:18

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For b) I think you are over-counting hands with more than one set of four.
  $endgroup$
  – Especially Lime
  Mar 19 at 12:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, do you think the question has restricted to only $1$? Anyway, thank you for giving me that feedback. I added the special case too...
  $endgroup$
  – Mostafa Ayaz
  Mar 19 at 12:23
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MostafaAyaz: No, that is not the problem. You are counting hands with e.g. four Threes and four Sevens twice. So your final answer is too large. (I think you should remove your edit $-$ it is not relevant.)
  $endgroup$
  – TonyK
  Mar 19 at 12:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. I don't understand. Why twice? First I count the number of hands containing at least one 4-tuple. This is possible in 13 different ways. After removing that 4-tuple, we are left with $48$ more cards from which we are supposed to choose the other $9$ cards of ours. This is possible in $binom489$. In case at least 2 four-tuples are to be drawn out, nothing is counted twice since I used $nCr$.
  $endgroup$
  – Mostafa Ayaz
  Mar 19 at 12:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MostafaAyaz the point is that one of the options your method counts is to remove all four threes, then choose nine more cards which are A,K,Q,J,10 of spades and all the sevens. Another option you've counted is to remove all four sevens, then choose nine more cards consisting of A,K,Q,J,10 of spades and four threes. But these two options are the same hand.
  $endgroup$
  – Especially Lime
  Mar 19 at 13:28
  

  
  
  
  

add a comment | 

$begingroup$

For $(a)$, the number of such hands can be calculated as follows:$$X=textnumber of all possible hands\-textnumber of all possible hands with no picture card\=binom5213-binom52-4times 413$$

For $(b)$, we choose a 4-tuple of same-rank cards in $13$ different ways, leaving the other cards to be chosen independently; totally$$Y=13timesbinom52-413-4$$

Edit

In case you want exactly 1 4-tuple, the answer is $$Y=13timesbinom489-binom132timesbinom445$$

For $(c)$, it is simply $$Z=4^13$$(why?)

share|cite|improve this answer

edited Mar 19 at 12:26

answered Mar 19 at 12:18

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

share|cite|improve this answer

edited Mar 19 at 12:26

answered Mar 19 at 12:18

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 19 at 12:18

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 19 at 12:18

Mostafa AyazMostafa Ayaz

18.2k31040

0

$begingroup$

For a) what you have is wrong on two counts: it is a probability, not a number of hands, and it is the probability of all cards being pictures.

The first point is easily fixed: just multiply by the total number of hands which is $binom5213$. The easiest way to correct the second point is to notice that this method can be used to calculate the probability of all cards being non-pictures: subtract that from 1 to get the probability of at least one picture.

share|cite|improve this answer

answered Mar 19 at 12:17

Especially LimeEspecially Lime

22.7k23059

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that for (a), he's calculating including ordering, meaning it's not $binom5213$.
  $endgroup$
  – Arthur
  Mar 19 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur The question just asks for the number of hands of this type, and that normally means different orderings count as the same hand.
  $endgroup$
  – Especially Lime
  Mar 19 at 12:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I agree that that's probably what the problem means. But that's not the OP's approach, as can be seen from the $frac1652times frac1551timescdots$ calculation, meaning your answer ishas not adressed all the issues with (a)
  $endgroup$
  – Arthur
  Mar 19 at 12:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur if you're calculating probabilities, it doesn't matter whether you take account of ordering or not, you get the same answer. So I don't think this is really an issue.
  $endgroup$
  – Especially Lime
  Mar 19 at 13:24
  

  
  
  
  

add a comment | 

0

$begingroup$

For a) what you have is wrong on two counts: it is a probability, not a number of hands, and it is the probability of all cards being pictures.

The first point is easily fixed: just multiply by the total number of hands which is $binom5213$. The easiest way to correct the second point is to notice that this method can be used to calculate the probability of all cards being non-pictures: subtract that from 1 to get the probability of at least one picture.

share|cite|improve this answer

answered Mar 19 at 12:17

Especially LimeEspecially Lime

22.7k23059

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that for (a), he's calculating including ordering, meaning it's not $binom5213$.
  $endgroup$
  – Arthur
  Mar 19 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur The question just asks for the number of hands of this type, and that normally means different orderings count as the same hand.
  $endgroup$
  – Especially Lime
  Mar 19 at 12:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I agree that that's probably what the problem means. But that's not the OP's approach, as can be seen from the $frac1652times frac1551timescdots$ calculation, meaning your answer ishas not adressed all the issues with (a)
  $endgroup$
  – Arthur
  Mar 19 at 12:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur if you're calculating probabilities, it doesn't matter whether you take account of ordering or not, you get the same answer. So I don't think this is really an issue.
  $endgroup$
  – Especially Lime
  Mar 19 at 13:24
  

  
  
  
  

add a comment | 

$begingroup$

For a) what you have is wrong on two counts: it is a probability, not a number of hands, and it is the probability of all cards being pictures.

The first point is easily fixed: just multiply by the total number of hands which is $binom5213$. The easiest way to correct the second point is to notice that this method can be used to calculate the probability of all cards being non-pictures: subtract that from 1 to get the probability of at least one picture.

share|cite|improve this answer

answered Mar 19 at 12:17

Especially LimeEspecially Lime

22.7k23059

$endgroup$

share|cite|improve this answer

answered Mar 19 at 12:17

Especially LimeEspecially Lime

22.7k23059

answered Mar 19 at 12:17

Especially LimeEspecially Lime

22.7k23059

answered Mar 19 at 12:17

Especially LimeEspecially Lime

22.7k23059