#$&*
course MTH 152
Time: 5:18 PM ET, Date: 7/8/13
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.
005. `query 5
question 11.5.10 ways to get different number on each of two fair dice In how many ways can you get a different number on each of two fair dice? How would you obtain the answer to this question if you were marooned on a desert island with no book and no table?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your Solution:
- First I define the number of ways one can get a different number.
- Each of the two die will have six sides (as a cube will have) so there are 12 different numbers in all that can be chosen. However, when each of the die are rolled, this will result in various combinations of those numbers.
- So by using the Fundamental Counting Principle one could hypothesize that there are 6*6 = 36 possible combinations in total.
- What if each of the die instead had letters, rather than numbers?
- Die 1 {a, b, c, d, e, f}
- Die 2 {g, h, i, j, k, l}
- This pattern would continue as (a f, g l) giving a total of 36 combinations.
- **** Note that there would be repetitions with this example, and thus we hit a constraint, how many DIFFERENT combinations****
- It follows that there are 6 combinations that repeat themselves (a,a) (b, b) (c, c) (d, d) (e, e) and (f, f)
- Thus the total minus the combinations that are the same
- = 36 - 6 = 30
- So therefore there are 30 combinations that can be rolled from two die.
*** I had initially started to go too far in my estimations. In other words, I second-guessed myself with using the FCP. Please provide guidance on when to use the counting principle and what the result will show.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:: ** On two fair dice you have 6 possible outcomes on the first and 6 possible outcomes on the second.
By the Fundamental Counting Principle there are therefore 6 * 6 = 36 possible outcomes.
We can list these outcomes in the form of ordered pairs:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Of these 36 outcomes there are six that have the same number on both dice. (i.e., (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6), running along the main diagonal of the table).
It follows that the remaining 3 - 6 = 30 have different numbers on the two dice.
So there are 30 ways to get different numbers on the two dice.
Your probability of getting different numbers when rolling two fair dice is therefore 30 / 36 = 5/6 = .8333... .**
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): Noted your comment to use discretion in creativity with mathematics. I tend to go too far and miss the point completely.
------------------------------------------------
Self-critique Rating: Ok. See my comment above in Self Critique and in Your Solution.
@&
Better to go too far and have to pull back a little than not to go anywhere at all.
*@
question 11.5.12 A bridge hand consists of 13 cards. A full deck of cards contains 52 cards, with 13 cards in each of the four different suits. How many bridge hands contain more than one suit?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your Solution:
- I’m unfamiliar with card games, so I look ahead to see what a suit entails.
- Made a note to read directions carefully. When re-writing the question for my notes, I noticed “ 13 cards in each of the four different suits” Which I assume 13 * 4 = 52, so there are 13 cards in a suit.
- My thought process begins at putting numbers to definitions.
- A bridge hand will have 13 cards, out of 52. So this is like taking 13 cards from an unordered collection of 52.
- C (52, 13) = 52! / [(52 - 13)! * 13!)]
- C (52, 13) = 52! / [ (37! * 13!)
- C (52, 13) = 52 * 51 * 50 * 49 * 48 * 47 * 46 * 45*44*43**42*41*40….
- This number, being so large, is irrelevant for now only to state that C(52, 13) = the number of possible combinations one can have when playing bridge.
- Otherwise, I think it’s noteworthy to state that I know absolutely nothing about cards; I need to be redirected with this question.
- Please provide a paraphrase of your Given Solution.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:: ** There are 13 cards in a bridge hand. The number of possible bridge hands is therefore C(52, 13).
There are 13 cards of each suit. The number of possible bridge hands with all cards in a given suit is therefore C(13, 13) = 1 (common sense is that there is only one way to get all 13 cards in a given suit, which is to get all the cards there are in that suit).
Since there are 4 suits there are 4 * C(13, 13) = 4 * 1 = 4 possible one-suit hands.
The number of hands having more than one suit is therefore C(52, 13) - 4. **
@&
There are C(52, 13) possible hands, as you indicate.
There is only one possible hand consisting of 13 cards of a given suit.
There are four suits.
So there are 4 possible one-suit hands.
*@
11.5.36 /6& 20 # subsets of 12-elt set with from 3 to 9 elts?
How many subsets contain from three to nine elements and how did you obtain your answer (answer in detail)?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your Solution:
- In a given 12 element set, say {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
- How many subsets contain from 3-9 elements?
- I consider that 9 - 3 = 6, so
C(12, 6) = 12! / [ (12 - 6)! * 6! = 12 * 11 * 10 * 9 * 8 * 7 / 6 * 5 * 4 * 3 * 2 * 1
- = 12/6 * 11/2 * 10/5 * 9/3 * 8/4 * 7/1 = 924 subsets.
- Therefore there are 924 subsets of 3 to 9 elements in a 12 element set.
- *********
- We can determine this either by listing out the number of possible subsets of C(12,3) through C(12, 9) or we can go the shorter route by determining how many subsets include the numbers that do not include 3 - 9 elements.
- That being said, I will take the cumulative sum of each subset C (12,1), C (12, 2), C(12, 10) , C( 12, 11) and C(12,12) subtracted from the maximum number of subsets.
------
Like I said before, I’m good with recognizing patterns. I love to look for them, and I get unnaturally excited when I find one.
First thing I do with this problem is list out my options. I could either
a - list out each possible subset whose elements list from 3 to 9 elements, take the sum.
B - list out each possible subset whose elements do NOT include 3 to 9 elements.
I take route A.
I list out as follows:
- C(12,3) = 220
- C(12,4) = 495
- C(12,5) = 792
- C(12,6) = 924
- C(12,7) = 792
- C(12, 8) = 495
- C(12, 9) = 220
- Whose sum of all the possible subsets in which the number of elements list from 3 to 9 = 3938.
** We can determine the number of subsets with 3 elements, with 4 elements, etc.. If we then add these numbers we get the total number of 3-, 4-, 5-, ., 9-element subsets.
Start with a 3-element subset. In a 12-element set, how many subsets have exactly three elements?
We can answer this by asking how many possibilities there are for the first element of our 3-element subset, then how many for the second, then how many for the third.
We can choose the first element from the entire set of 12, so we have 12 choices.
We have 11 elements remaining from which to choose the second, so there are 11 choices.
We then have 10 elements left from which to choose the third.
So there are 12 * 11 * 10 ways to choose the three elements of our subset if we choose them in order.
However, the order of a set doesn't matter. The set is the same no matter in which order its elements have been selected.
The three elements of the subset could be ordered in 3! different ways, so there are 12 * 11 * 10 / 3! ways to choose different three-element sets.
12 * 11 * 10 / 3! is the same as C(12,3).
So there are C(12, 3) three-element subsets of a set of 12 elements.
Reasoning similarly we find that there are
C(12,4) ways to choose a 4-element subset.
C(12,5) ways to choose a 5-element subset.
C(12,6) ways to choose a 6-element subset.
C(12,7) ways to choose a 7-element subset.
C(12,8) ways to choose a 8-element subset.
C(12,9) ways to choose a 9-element subset.
Thus there are
C(12,3) + C(12,4) + C(12,5) + C(12,6) + C(12,7) + C(12,8) + C(12,9)
possible subsets with 3, 4, 5, 6, 7, 8 or 9 elements.
Alternative (and shorter) solution:
It is easier to figure out how many sets have fewer than 3 or more than 9 elements.
There are C(12, 0) + C(12, 1) + C(12, 2) = 1 + 12 + 66 = 79 sets with fewer than 3 elements, and C(12, 10) + C(12, 11) + C(12, 12) = 66 + 12 + 1 = 79 sets with more than 3.
Since there are 2^12 = 4096 possible subsets of a 12-element set there are 4096 - 79 - 79 = 3938 sets with between 3 and 9 elements. **
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): Is there a relationship between the number of elements, ‘n’ in comparison with 2^n? If there is, wouldn’t that mean that, for any set containing ‘n’ number of elements, there is an exponential relationship to the number 2? --- Say we have a set containing 13 elements. Would that dictate the maximum number of subsets being 2^13?
@&
Very good.
There are 2^n possible subsets of a set containing n elements.
*@
*********************************************
Question: 11.5.&47 / 30 10200 ways to get a straight Verify that there are in fact 10200 ways to get a straight in a 5-card hand.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your Solution:
- My first guess was to do the following
- C(52, 5) = not the droids I’m looking for; this instead just gave me the number of possible ways five cards can be chosen at random from a deck of 52.
- A straight in cards is, by definition, apparently 5 cards in sequence of the same suit, such as
- Q, J, 10, 9, 8.
- My second attempt suggests considering four suits per deck with 13 cards in each suit.
- So I hypothesize taking C(13,5) * 4, but my answer yielded 5,148. Not 10,200 like the question states.
- Maybe I need to consider the suit of cards as a set of elements, like
- (K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3, 2, A)
- In which I get a hand such as
- (K, Q, J, 10, 9)
- (Q, J, 10, 9, 8)
- (J, 10, 9, 8, 7)
- (10, 9, 8, 7, 6)
- (9, 8, 7, 6, 5)
- (8, 7, 6, 5, 4)
- (7, 6, 5, 4, 3)
- (6, 5, 4, 3, 2)
- (5, 4, 3, 2, A)
- In which there are 9 ways to get a 5-card straight, working under the assumption that a 5-card straight is a combination in which each card is exactly one ‘power’ lower than the last.
- So this would mean there are 9 ways taken from 13, so
C(13, 9) = 13! / [ (13 - 9) ! * 9!] = 13 * 12 *11 * 10 * 9 * 8 * 7 * 6 * 5 /9 * 8 * 7 * 6 *5 *4 *3 *2 *1 = 715….Still off.
- After the second attempt I glance ahead to see what direction you had taken in Given Solution:
-
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:: ** A ‘straight’ consists of five consecutive cards in the same suit (for example 4-5-6-7-8 of diamonds, or 8-9-10-Jack-Queen of hearts).
The highest possible straight is therefore 10, Jack, Queen, King and Ace.
We enumerate the possible ‘straights’ by counting the number of choices for each card in turn, starting with the lowest and working up to the highest.
The lowest straight would be 2-3-4-5-6. In some games the Ace can count as the high card or the low card, in which case the lowest straight would be ace-2-3-4-5.
The lowest card of a straight can therefore be any of the nine numbers from 2 through 10, plus the Ace (provided it can be counted as the low card), which makes 10 possible low cards.
With each denomination appearing in four suits there are thus 4 * 10 = 40 cards that could be the lowest card of a straight.
There are then four choices for the next-higher card, four for the next after that, etc., giving us 40*4*4*4*4 possibilities for the five cards.
This gives us 40 * 4 * 4 * 4 * 4 = 10240 possible ‘straights’. The specified answer is 10200. The difference occurs because 40 of the straights are really ‘straight flushes’. A ‘straight’ is a good hand, beating most other hands; however there are hands that beat a ‘straight’, and of all hands the ‘straight flush’ is the very best. The only hand that can beat a ‘straight flush’ is a higher ‘straight flush’:
A ‘straight flush’ is a straight in which all the cards are of the same suit.
Recall that there are 40 possible lowest cards for a straight. To get a ‘straight flush’, once the lowest card is chosen, every subsequent card must match its suit. There is thus only one choice for each of the remaining cards (the second card is the next-higher card of the same suit; the third is the next-higher card of that same suit; etc.).
So the number of ‘straight flushes’ is 40 * 1 * 1 * 1 * 1 = 40, and the number of ‘just plain straights’ becomes 10240 - 40 = 10200.
We could actually list the ‘straights’:
Ace-2-3-4-5, where there are 4 choices for the Ace, four for the 2, four for the 3, four for the 4 and four for the 5, giving us 4 * 4 * 4 * 4 * 4 = 1024 ways to get a ‘straight’ in this hand; however there are four ‘straight flushes’, one for each suite running from Ace through 5 (i.e., Ace-2-3-4-5 of Hearts, Ace-2-3-4-5 of Diamonds, Ace-2-3-4-5 of clubs, Ace-2-3-4-5 of Spades), and this reduces the number to 1020.
2-3-4-5-6, where again there are four choices for the suit of each card, yielding 4^5 = 1024 possibilities, of which 4 would be straight flushes and would not be included.
3-4-5-6-7, where again there are 1024 - 4 = 1020 ways to get straight which is not a straight flush.
4-5-6-7-8, where once more the number of possibilities is 1020.
… etc … up to
10-Jack-Queen-King-Ace, with an additional 1020 possibilities.
Our listing would consist of 10 * 1020 = 10200 possible choices.
STUDENT QUESTION: Where did you get the 40*4*4*4*4? The four'S were confusing to me?
INSTRUCTOR RESPONSE: Any of 40 cards can be the lowest in a 'straight'. Any card except a Jack, Queen or King can be the lowest card in a straight.
The given solution included the reasoning (repeated from a previous qa) for the number of regular 'straights'; this isn't necessary to answer the present question, but it provides a useful contrast to the present solution. To get a regular 'straight' the next-higher card would be one of the four cards of the next-higher denomination (e.g., if the lowest card was the 5 of diamonds, the next-higher card could be any of the four 6's). So there are then four choices for the next-higher card, four for the next card after that, etc..
To get a 'straight flush' there is only one choice for the next card, which must match the suit of the first card, and for similar reason there is only one choice for each of the remaining cards. For example, if the lowest was the 5 of diamonds, the remaining cards would be the 6, 7, 8 and 9, all of diamonds.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): Didn’t understand the parameters of the question because I wasn’t aware that an Ace could be counted as a high/low card, but I can see that this would alter the subject matter; but not HOW it would be changed. Is there any way you can paraphrase this question so that I may do a problem similar to this one? Or any way you could help cement my understanding?
I can see that the J, K, Q would be exempt from being a ‘low’ card, mathematically speaking this is 13 -3 = 10, then 10 * 4 suits = 40 combinations of cards that can be the lowest in a hand? From there I see that you take 40 * 4 * 4 *4 * 4, which I figured would be read as “40 possible combinations of cards that can be the lowest in a hand, multiplied by the number of suits total, 4, four times.”
Please provide any insight or clarity. I do like where this is going, though!
------------------------------------------------
Self-critique Rating: Ok, see comment above.
@&
Good.
There are 40 cards that can be the lowest in a straight.
Given the low card:
There are 4 cards one denomination above it, 4 cards two denominations above the low card, 4 cards three denominations above the low card, 4 cards four denominations above the low card.
By the Fundamental Counting Principles, then, there are 4 * 4 * 4 * 4 hands with the given low card.
Multiply this by the 40 possible low cards and you have the result.
*@
*********************************************
question: Note that the 11th edition of the text has unfortunately eliminated the subsequent problems from Section 11.5:
11.5.36 xxxxxxxxxxxxxx 3-digit #'s from {0, 1, ..., 6}; how many mult of 25?
** A 3-digit number from the set has six choices for the first digit (can't start with 0) and 7 choices for each remaining digit. That makes 6 * 7 * 7 = 294 possibilities.
A multiple of 25 is any number that ends with 00, 25, 50 or 75. SInce 7 isn't in the set you can't have 75, so there are three possibilities for the last two digits. There are six possible first digits, so from this set there are 6 * 3 = 18 possible 3-digit numbers which are multiples of 25.
A listing would include 100, 125, 150, 200, 225, 250, 300, 325, 350, 400, 425, 450, 500, 525, 550, 600, 625, 650.
Combinations aren't appropriate for two reasons. In the first place the uniformity criterion is not satisfied because different digits have different criteria (i.e., the first digit cannot be zero). In the second place we are not choosing object without replacement.
The fundamental counting principle is the key here.
STUDENT SOLUTION AND INSTRUCTOR RESPONSE: All I can come up with is C(7,2)=21. & choices of #s and the # must end in 0 or 5 making it 2 of the 7 choices
INSTRUCTOR RESPONSE: Right reasoning on the individual coices but you're not choosing just any 3 of the 7 numbers (uniformity criterion isn't satisfied--second number has different criterion than first--so you wouldn't use permutations or combinations) and order does matter in any case so you wouldn't use combinations.
You have 7 choices for the first and 2 for the second number so there are 7 * 2 = 14 multiples of 5. **
Query 11.5.48 xxxxxxxxxxxxxxx # 3-digit counting #'s without digits 2,5,7,8?
** there are 5 possible first digits (1, 3, 4, 6, or 9) and 6 possibilities for each of the last two digits. This gives you a total of 5 * 6 * 6 = 180 possibilities. **
Query Add comments on any surprises or insights you experienced as a result of this assignment.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): See my comment with regards to verifying there are 10200 ways to get a straight hand in a game of cards. I think that this can be very handy to commit to memory.
------------------------------------------------
Self-critique Rating:
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
&#Your work looks good. See my notes. Let me know if you have any questions. &#