#$&* course MTH 152 2/20 7 005. `query 5question 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?
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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 36 - 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): OK ------------------------------------------------ Self-critique Rating: OK 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: There are 52 cards in a full deck, which means that there are only so many bridge hands that can be created from a deck of 52 cards. The number of possible bridge hands from a full deck of cards can be written as C(52,13), and because there are only four possible times that a full suit could be contained within a bridge hand (as both a suit of cards and a bridge deck have 13 cards), then we must account for that limited number of times. Our final equation used to find out how many bridge hands contain more than one suit is C(52,13)-4. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** Question 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: If we have a set of 12 elements, and we are trying to discover how many subsets include 3 elements, 4 elements, 5 elements, 6 elements, 7 elements, 8 elements, and 9 elements, we can use the same sort of equation to solve each problem. In order to discover how many subsets of a 12 element set will contain 3 elements, we must consider that there are twelve choices from which to choose the three elements, which gives us 12*11*10 ways to choose the three element subsets. This gives us an equation of 12*11*10, and accounting for the fact that order does not matter, a final equation of 12*11*10/3*2*1, or C(12,3). We use the same exact method in order to find out how many subsets include 4 elements, 5 elements, 6 elements, 7 elements, 8 elements, and 9 elements: For 4 elements we have an equation of 12*11*10*9/4!, or C(12,4). For 5 elements we have an equation of 12*11*10*9*8/5!, or C(12,5). For 6 elements we have an equation of 12*11*10*9*8*7/6! or C(12,6). For 7 elements we have an equation of 12*11*10*9*8*7*6/7!, or C(12,7). For 8 elements we have an equation of 12*11*10*9*8*7*6*5/8!, or C(12,8). For 9 elements we have an equation of 12*11*10*9*8*7*6*5*4/9!, or C(12,9). When we add all of these values [C(12,3), C(12,4), C(12,5), C(12,6), C(12,7), C(12,8), C(12,9)] together we discover how many subsets contain 3 elements, 4 elements, 5 elements, 6 elements, 7 elements, 8 elements, and 9 elements. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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): OK ********************************************* 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: There are 52 cards in any fair deck of cards and 40 cards that are not a Jack, Queen, or King. Any of the 40 remaining cards could be the beginning of a straight. This accounts for our first card in our 5-card hand. There are 4 suits of cards in a deck, and always 4 other possibilities of drawing a higher card than our first card. This gives us an equation of 40*4*4*4*4. 40*4*4*4*4 is 10,200, which verifies that there are indeed 10,200 ways to get a straight in a 5-card hand. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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 choices 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): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!