assignment 6 qa

#$&*

course Mth 152

02/02/2014, 4:20 p.m.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution.

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

006. Cards

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Question: `q001. Note that there are 10 questions in this assignment.

A standard deck of cards consists of four suits, each containing 13 cards. Ten of the 13 cards in each suit are numbered 1 - 10, and three are called the Jack, Queen and King. The card numbered 1 is also often referred to as an 'ace'. The number or label of a card is its denomination (so the denominations are 1 (or 'ace'), 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen and King.

The four suits are hearts, diamonds, clubs and spades.

The hearts and diamonds are red, clubs and spades are black.

Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

c[4,2]

4!/2![4-2!]

4*3/2=6

52-4=48

5-2=3

c[48,3]

48!/3![48-3]!

48*47*46/6=103776/6=17296

17296*6=103776

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: In order to get a hand containing exactly two 5's we must select, without regard for order, two of the four 5's, then we must select the remaining 3 cards from the 48 cards that are not 5's.

There are C(4,2) ways to select two 5's from the four 5's in the deck.

There are C(48,3) ways to select the 3 remaining cards from the 48 cards which are not 5's.

We must do both, so by the Fundamental Counting Principle there are C(4,2) * C(48, 3) ways to obtain exactly two 5's.

Self-critique:ok

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Self-critique rating:ok

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Question: `q002. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's and exactly two 9's?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

C[4,2]*C[4,2]*C[44,1]

C[4,2}

4!/2![4-2]!=12/2=6

52-8=44

5-4=1

c[44,1]=44

6*6*44=1584

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: There are C( 4, 2) ways to select exactly two 5's and C(4, 2) ways to select exactly two 9's. There are 44 remaining cards which are neither 9 nor 5.

The total number of possible ways is therefore C(4, 2) * C(4, 2) * 44.

Self-critique:OK

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Self-critique rating:OK

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Question: `q003. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three 9's?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

C[4,2]*C[4,3]

4!/2![4-2]!

12/2=6

4!/3![4-3]!

4*3/6=2

6*2=12

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: There are C(4, 2) ways to get two 5's and C(4, 3) ways to get three 9's. It follows that the number of ways to get the specified 'full house' with two 5's and three 9's is C(4,2) * C(4,3).

Self-critique:OK

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Self-critique rating:OK

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Question: `q004. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three identical face cards?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

full house

C[4,2]=6

c[4,3]=2

6*2=12

full house with three identical face cards

6*2*3=36

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: There are C(4,2) ways to get two 5's and C(4,3) ways to get three of any given face card. There are 3 possible face cards, so the number of ways to get a 'full house' consisting of two 5's and three identical face cards is 3 * C(4,2) * C(4,3).

Self-critique:ok

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Self-critique rating:ok

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Question: `q005. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two of one denomination and three of another?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

C[4,2]*C[4,3]

4!/2![4-2]!

12/2=6

4!/3![4-3]!

4*3/6=2

6*2=12

13 possible pairs, 12 three of a kind

13*12*12=1872

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: For any two denominations, e.g., a pair of 5's and three 9's, there are C(4,2) * C(4,3) different full houses.

There are 13 possible choices for the pair, which leaves 12 possible choices for the three-of-a-kind, which to by the Fundamental Counting Principle makes 13 * 12 possible choices for the two denominations. Note that order does matter when choosing the denominations because one has to be the pair and the other the three-of-a-kind.

Again by the Fundamental Counting Principle we conclude that there are 13 * 12 * C(4,2) * C(4,3) possible full houses.

Self-critique:ok

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Self-critique rating:ok

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Question: `q006. Using a standard deck of cards, in how many ways is it possible to get a 'flush' consisting of five cards all of the same suit?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

c[13,5]

13!/5![13-5]!

13*12*11*10*9/5*4*3*2*1=154440/120=1287

4 suits

5148

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: There are 13 cards in any given suit, so there are C(13,5) hands consisting of all 5 cards in that suit. There are 4 suits, so there are 4 * C(13,5) possible flushes.

Self-critique:ok

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Self-critique rating:ok

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Question: `q007. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of exactly one each of the denominations 5, 6, 7, 8 and 9?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

there are four of each of these denominations, c[4,1]=4

c[4,1]*c{4,1]*c[4,1]*c[4,1]*c[4,1]

4^5=4*4*4*4*4=1024

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: There are four 5's, four 6's, four 7's, four 8's and four 9's. So there are 4 * 4 * 4 * 4 * 4 = 4^5 possible straights from 5 to 9.

STUDENT QUESTION

not sure I understand why is it not C(20,5)

I bet because it is C(4,1)*C(4,1)…..or 4^5 or maybe not…

INSTRUCTOR RESPONSE

There are indeed 20 cards which are 5, 6, 7, 8 or 9, which four of each of the five denominations.

However of the C(20, 5) combinations of 5 of the 20 cards, only a few have one of each denomination. For example, one of the C(20, 5) combinations would be 5 of hearts, 5 of spades, 5 of diamonds, 7 of clubs and 9 of hearts.

That's not a straight, nor are most of the C(20, 5) combinations of these cards.

C(4,1) * C(4,1) * C(4,1) * C(4,1) * C(4,1) = 4*4*4*4*4 = 4^5, so that is correct. You can check to see that this is a good bit less than C(20, 5).<

Self-critique:ok

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Self-critique rating:ok

*********************************************

Question: `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

4^5 can be used from the previos problem, 10 possible denomination cards

10*4^5=10240

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: There are 10 possible denominations for the 'low' card of a straight (any card from ace through 10 can serve as the 'low' card of a straight). There are 4^5 possible straights for each possible low card. It follows that there are 10 * 4^5 possible straights.

STUDENT QUESTION: Where does the 4^5 come from? I understand the 10 and 4 different cards of each but did you get the ^5 from the problem before this one?

INSTRUCTOR RESPONSE 4^5 is the number of possibilities for each denomination of the 'low' card. For example we figured out above that there are 4^5 straights with 5 as the low card.

Similarly there are 4^5 straights with the 'ace' low, another 4^5 with the 2 low, etc., with the 'highest' possible 'low card being 10. So there are 10 possible denominations for the 'low' card, and 4^5 possible straights for each 'low' denomination.

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Question: `q009. Using a standard deck, in how many ways is it possible to get a 5-card hand consisting of all face cards?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

c[16,5]

16!/5![16-5]!

16*15*14/120=28

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q010. Using a standard deck, in how many ways is it possible to get a 5-card hand which includes exactly two face cards?

(Optional challenge question: What is more probable, a 5-card hand consisting of exactly two face cards, or a 5-card hand consisting of no face cards?)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

c[16,2]*c[5,3]

52-16=36

5-2=3

c[16,2]

16*15/2=120

c[5,3]

5*4*3/6=10

120*10=1200

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Self-critique Rating:ok

"

Self-critique (if necessary):

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Self-critique rating:

&#Very good responses. Let me know if you have questions. &#

assignment 6 qa

#$&*

course Mth 152

02/02/2014, 4:20 p.m.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution.

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

006. Cards

*********************************************

Question: `q001. Note that there are 10 questions in this assignment.

A standard deck of cards consists of four suits, each containing 13 cards. Ten of the 13 cards in each suit are numbered 1 - 10, and three are called the Jack, Queen and King. The card numbered 1 is also often referred to as an 'ace'. The number or label of a card is its denomination (so the denominations are 1 (or 'ace'), 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen and King.

The four suits are hearts, diamonds, clubs and spades.

The hearts and diamonds are red, clubs and spades are black.

Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

c[4,2]

4!/2![4-2!]

4*3/2=6

52-4=48

5-2=3

c[48,3]

48!/3![48-3]!

48*47*46/6=103776/6=17296

17296*6=103776

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: In order to get a hand containing exactly two 5's we must select, without regard for order, two of the four 5's, then we must select the remaining 3 cards from the 48 cards that are not 5's.

There are C(4,2) ways to select two 5's from the four 5's in the deck.

There are C(48,3) ways to select the 3 remaining cards from the 48 cards which are not 5's.

We must do both, so by the Fundamental Counting Principle there are C(4,2) * C(48, 3) ways to obtain exactly two 5's.

Self-critique:ok

------------------------------------------------

Self-critique rating:ok

*********************************************

Question: `q002. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's and exactly two 9's?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

C[4,2]*C[4,2]*C[44,1]

C[4,2}

4!/2![4-2]!=12/2=6

52-8=44

5-4=1

c[44,1]=44

6*6*44=1584

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: There are C( 4, 2) ways to select exactly two 5's and C(4, 2) ways to select exactly two 9's. There are 44 remaining cards which are neither 9 nor 5.

The total number of possible ways is therefore C(4, 2) * C(4, 2) * 44.

Self-critique:OK

------------------------------------------------

Self-critique rating:OK

*********************************************

Question: `q003. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three 9's?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

C[4,2]*C[4,3]

4!/2![4-2]!

12/2=6

4!/3![4-3]!

4*3/6=2

6*2=12

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: There are C(4, 2) ways to get two 5's and C(4, 3) ways to get three 9's. It follows that the number of ways to get the specified 'full house' with two 5's and three 9's is C(4,2) * C(4,3).

Self-critique:OK

------------------------------------------------

Self-critique rating:OK

*********************************************

Question: `q004. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three identical face cards?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

full house

C[4,2]=6

c[4,3]=2

6*2=12

full house with three identical face cards

6*2*3=36

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: There are C(4,2) ways to get two 5's and C(4,3) ways to get three of any given face card. There are 3 possible face cards, so the number of ways to get a 'full house' consisting of two 5's and three identical face cards is 3 * C(4,2) * C(4,3).

Self-critique:ok

------------------------------------------------

Self-critique rating:ok

*********************************************

Question: `q005. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two of one denomination and three of another?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

C[4,2]*C[4,3]

4!/2![4-2]!

12/2=6

4!/3![4-3]!

4*3/6=2

6*2=12

13 possible pairs, 12 three of a kind

13*12*12=1872

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: For any two denominations, e.g., a pair of 5's and three 9's, there are C(4,2) * C(4,3) different full houses.

There are 13 possible choices for the pair, which leaves 12 possible choices for the three-of-a-kind, which to by the Fundamental Counting Principle makes 13 * 12 possible choices for the two denominations. Note that order does matter when choosing the denominations because one has to be the pair and the other the three-of-a-kind.

Again by the Fundamental Counting Principle we conclude that there are 13 * 12 * C(4,2) * C(4,3) possible full houses.

Self-critique:ok

------------------------------------------------

Self-critique rating:ok

*********************************************

Question: `q006. Using a standard deck of cards, in how many ways is it possible to get a 'flush' consisting of five cards all of the same suit?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

c[13,5]

13!/5![13-5]!

13*12*11*10*9/5*4*3*2*1=154440/120=1287

4 suits

5148

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: There are 13 cards in any given suit, so there are C(13,5) hands consisting of all 5 cards in that suit. There are 4 suits, so there are 4 * C(13,5) possible flushes.

Self-critique:ok

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Self-critique rating:ok

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Question: `q007. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of exactly one each of the denominations 5, 6, 7, 8 and 9?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

there are four of each of these denominations, c[4,1]=4

c[4,1]*c{4,1]*c[4,1]*c[4,1]*c[4,1]

4^5=4*4*4*4*4=1024

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: There are four 5's, four 6's, four 7's, four 8's and four 9's. So there are 4 * 4 * 4 * 4 * 4 = 4^5 possible straights from 5 to 9.

STUDENT QUESTION

not sure I understand why is it not C(20,5)

I bet because it is C(4,1)*C(4,1)…..or 4^5 or maybe not…

INSTRUCTOR RESPONSE

There are indeed 20 cards which are 5, 6, 7, 8 or 9, which four of each of the five denominations.

However of the C(20, 5) combinations of 5 of the 20 cards, only a few have one of each denomination. For example, one of the C(20, 5) combinations would be 5 of hearts, 5 of spades, 5 of diamonds, 7 of clubs and 9 of hearts.

That's not a straight, nor are most of the C(20, 5) combinations of these cards.

C(4,1) * C(4,1) * C(4,1) * C(4,1) * C(4,1) = 4*4*4*4*4 = 4^5, so that is correct. You can check to see that this is a good bit less than C(20, 5).<

Self-critique:ok

------------------------------------------------

Self-critique rating:ok

*********************************************

Question: `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

4^5 can be used from the previos problem, 10 possible denomination cards

10*4^5=10240

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: There are 10 possible denominations for the 'low' card of a straight (any card from ace through 10 can serve as the 'low' card of a straight). There are 4^5 possible straights for each possible low card. It follows that there are 10 * 4^5 possible straights.

STUDENT QUESTION: Where does the 4^5 come from? I understand the 10 and 4 different cards of each but did you get the ^5 from the problem before this one?

INSTRUCTOR RESPONSE 4^5 is the number of possibilities for each denomination of the 'low' card. For example we figured out above that there are 4^5 straights with 5 as the low card.

Similarly there are 4^5 straights with the 'ace' low, another 4^5 with the 2 low, etc., with the 'highest' possible 'low card being 10. So there are 10 possible denominations for the 'low' card, and 4^5 possible straights for each 'low' denomination.

*********************************************

Question: `q009. Using a standard deck, in how many ways is it possible to get a 5-card hand consisting of all face cards?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

c[16,5]

16!/5![16-5]!

16*15*14/120=28

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q010. Using a standard deck, in how many ways is it possible to get a 5-card hand which includes exactly two face cards?

(Optional challenge question: What is more probable, a 5-card hand consisting of exactly two face cards, or a 5-card hand consisting of no face cards?)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

c[16,2]*c[5,3]

52-16=36

5-2=3

c[16,2]

16*15/2=120

c[5,3]

5*4*3/6=10

120*10=1200

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Self-critique Rating:ok