course 152
This assigment was hard because I do not know the
cards
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
Self-critique:
Self-critique rating:
Question: **** `q001. Note that there are 8 questions in this assignment.
Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's?
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Your solution:
(4,2)
52-4=48
(48,3)
There are 5 cards in a hand so you want the two to be 5’s and the other 3 cards will not be fives
C(4,2)*C(48,3)
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:
Yale has a hard time understanding what a deck of cards are?
She does very good on the problems but it’s the communication barrier that hurts her.
Self-critique rating:1.75
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?
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Your solution:
5’s= (4,2)
9’s= (4,2)
5 cards to a hand
4+4=8
52-8=44
(44,1)
2+2=4
5-4=1
C(4,2)*C(4,2)*C(44,1)
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:
This one made more sense to Yale
-Katye
Self-critique rating:
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?
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Your solution:
(4,2)=5’s
(4,3)=9’s
C(4,2)*C(4,3)
Confidence Rating:
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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:
Getting better at understanding a deck of cards…we found a picture on the internet
Self-critique rating:2.5
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?
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Your solution:
(4,2)=5’s
(4,3)=Jack’s
(4,3)=Queen’s
(4,3)=King’s
C(4,2)*C(4,3)*C(4,3)*C(4,3)
3*C(4,2)*C(4,3)
Confidence Rating:
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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:
Actually done this one with cards
Self-critique rating:3
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?
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Your solution:
C(4,2)*C(4,3)
P=n!/(n-r)!
4!/(4-2)!
4*3*2*1/2*1 4*3*2*1/1
4*3=12 4*3*2=24
21*24
Confidence Rating:
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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:
Where do you get the 13 from? Did I use the right formula?
I understand the C(4.2)*C(4,3)
But not the 13??
Did you get the 13 from the fact that there are 13 cards of each suit??
Yes. A pair consists of 2 cards of the same denomination (like two 3's, or two Queens). There are 13 denominations, with 4 suits in each denomination.
Self-critique rating:1
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?
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Your solution:
4 of each suit
(13,5) possible
4*C(13,5)
Confidence Rating: 2
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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:
Self-critique rating:
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?
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Your solution:
4 -5’s
4-6s
4-7s
4-8s
4-9s
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.
Self-critique:
Self-critique rating:2.5
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'?
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Your solution:
4*4*4*4*4*4*4*4*4*4
4^10 possible ways
Confidence Rating: 1
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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.
Self-Critique
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?
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.
&#This looks good. See my notes. Let me know if you have any questions. &#