#$&* course Mth 152 Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: When choosing 4 objects out of 12, there are 12 choices for the first, 11 choices for the second, 10 choices for the third and 9 choices for the fourth object. If the order matters there are therefore 12 * 11 * 10 * 9 possible outcomes. If the order doesn't matter, then we have to ask in how many different orders any given collection of 4 objects could be chosen. Given any 4 objects, there are 4 choices for the first, 3 choices for the second, 2 choices for the third and 1 choice for the fourth. There are thus 4 * 3 * 2 * 1 orders in which a given set of 4 objects could be chosen. We therefore have 12 * 11 * 10 * 9 / ( 4 * 3 * 2 * 1) possible outcomes when order doesn't matter. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. If order does not matter, then how many ways are there to choose 5 members of a team from 23 potential players? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 33649??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If order did matter then there would be 23 * 22 * 21 * 20 * 19 ways choose the five members. However order does not matter, so we must divide this number by the 5 * 4 * 3 * 2 * 1 ways in which any given set of five individuals can be chosen. We therefore have 23 * 22 * 21 * 20 * 19 / ( 5 * 4 * 3 * 2 * 1) possible 5-member teams. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. In how many ways can we line up 5 different books on a shelf? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 120??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: It should be clear that there are 5 * 4 * 3 * 2 * 1 ways, since there are 5 choices for the first book, 4 for the second, etc.. If we multiply these numbers out we get 5 * 4 * 3 * 2 * 1 = 120. It might be a little bit surprising that there should be 120 ways to order only 5 objects. It’s probably even more surprising that if we double the number of objects to 10, there are over 3 million ways to order them (you should be able to verify this easily enough). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. The expression 5 * 4 * 3 * 2 * 1 is often written as 5 ! , read 'five factorial'. More generally if n stands for any number, then n ! stands for the number of ways in which n distinct objects could be lined up. Find 6 ! , 7 ! and 10 ! . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 720??? 5040??? 3628800??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 6 ! = 6 * 5 * 4 * 3 * 2 * 1 = 720. 7 ! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040. 10 ! = 3,628,800. These numbers grow at an astonishing rate. The last result here shows is that there are over 3 million ways to arrange 10 people in a line. The rapid growth of these results like in part explain the use of the ! symbol (the exclamation point) to designate factorials. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. What do we get if we simplify the expression (10 ! / 6 !) ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5040??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 10 ! / 6 ! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 6 * 5 * 4 * 3 * 2 * 1). We can simplify this by rewriting it as 10 * 9 * 8 * 7 * (6 * 5 * 4 * 3 * 2 * 1) / ( 6 * 5 * 4 * 3 * 2 * 1) = 10 * 9 * 8 * 7. We see that the 6 * 5 * 4 * 3 * 2 * 1 in the numerator matches the same expression in the denominator, so when divided these expressions give us 1 and we end up with just 10 * 9 * 8 * 7 * 1 = 10 * 9 * 8 * 7. Note that this is just the number of ways in which 4 objects can be chosen, in order, from a collection of 10 objects. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. We saw above that there are 23 * 22 * 21 * 20 * 19 ways to choose 5 individuals, in order, from 23 potential members. How could we express this number as a quotient of two factorials? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4037880??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we divide 23 ! by 18 ! , the numbers from 18 down to 1 will occur in both the numerator and denominator and when we divide we will be left with just the numbers from 23 down to 19. Thus 23 * 22 * 21 * 20 * 19 = 23 ! / 18 !. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. How could we express the number of ways to rank 20 individuals, in order, from among 100 candidates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 100 ! / 80 !??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 100 choices for the first candidate, 99 for the second, 98 for the third, etc.. For the 20th candidate there are 81 choices. You should convince yourself of this if you didn't see it originally. Our product is therefore 100 * 99 * 98 * ... * 81, which can be expressed as 100 ! / 80 !. Note that the denominator is 80 !, which can be written as (100 - 20)! . So the result for this problem can be written as 100 ! / (100 - 20) ! = 100 ! / 80 ! = 100 * 99 * 98 * … * 81. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. How could we express the number of ways to rank r individuals from a collection of n candidates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: n ! / ( n - r ) ! ??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: By analogy with the preceding example, where we divided 100 ! by (100 - 20) !, we should divide n ! by ( n - r ) !. The number is therefore n ! / ( n - r ) !. This is the number of ways in which we can choose, in order, r objects from a collection of n objects. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q009. The expression n ! / ( n - r ) ! denotes the number of ways in which r objects can be chosen, in order, from among n objects. When we choose objects in order we say that we are 'permuting' the objects. The expression n ! / ( n - r ) ! is therefore said to be the number of permutations of r objects chosen from n possible objects. We use the notation P ( n , r ) to denote this number. Thus P(n, r) = n ! / ( n - r ) ! . Find P ( 8, 3) and explain what this number means. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 336??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: P(n, r) = n! / ( n - r) !. To calculate P(8, 3) we let n = 8 and r = 3. We get P(8, 3) = 8 ! / ( 8 - 3) ! = 8 ! / 5 ! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 5 * 4 * 3 * 2 * 1) = 8 * 7 * 6. This number represents the number of ways in which 3 objects can be chosen, in order, from 8 objects. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q010. In how many ways can an unordered collection of 3 objects be chosen from 8 candidates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 56??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 8 * 7 * 6 ways to choose 3 objects from 8, in order, and 3! ways to order any unordered collection of 3 objects, so there are 8 * 7 * 6 / ( 3 * 2 * 1 ) possible unordered collections. This number is easily enough calculated. Since 3 goes into 6 twice and 2 goes into 8 four times, we see that 8 * 7 * 6 / ( 3 * 2 * 1) = 4 * 7 * 2 = 56. There are 56 different unordered collections of 3 objects chosen from 8. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q010. How could the result of the preceding problem be expressed purely in terms of factorials? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8 ! / ( 5 ! * 3 ! ) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The product 8 * 7 * 6 is just 8 ! / 5 !, and the expression 8 * 7 * 6 / ( 3 * 2 * 1) can therefore be expressed as 8 ! / ( 5 ! * 3 !). QUESTION FROM STUDENT: How do you know to use 8!/ (5! * 3!) INSTRUCTOR'S ANSWER: The preceding problem involved choosing 3 objects out of 8. There would be 8 choices for the first item, 7 choices for the second and 6 choices for the third. If chosen in order, then by the fundamental counting principle there would be 8 * 7 * 6 possible choices. 8 * 7 * 6 = 8 ! / 5 ! , as can easily be seen by writing the factorials out: · 8 ! / 5 ! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / (5 * 4 * 3 * 2 * 1). · The (5 * 4 * 3 * 2 * 1) in the denominator is matched by the 5 * 4 * 3 * 2 * 1 in the numerator so these factors divide out, leaving just 8 * 7 * 6. Since we are choosing 3 objects out of 8. We want to write our result in terms of the numbers 3 and 8. Where does the number 5 in our expression 8 ! / 5 ! come from? · The answer is that to get 8 * 7 * 6 we need to 'chop off' the last 5 factors in 8 ! . This is why we divide by 5 !. · Since 8 ! contains 8 factors and we need to leave only the first three, we have to 'chop off' 8 - 3 = 5 of them. · Thus we divide by (8 - 3) ! , i.e., by 5 !. So our number of ordered choices can be expressed in three possible ways: · 8 * 7 * 6, which we get by applying the fundamental counting principle, · 8 ! / 5 !, which 'chops off' the last 5 factors of 8 !, leaving us 8 * 7 * 6, or · 8 ! / (8 - 3) !, which is how we write the result in terms of the original numbers 3 and 8. Thus the number of ordered choices is 8 * 7 * 6, or 8 ! / 5 !, or 8 ! / (8 - 3) ! · This number is denoted P(8, 3), the number of permutations (i.e., ordered choices) of 3 objects chosen without replacement from 8. · P(8, 3) = 8 ! / (8 - 3) !, and this is our official definition of P(8, 3). Working from this definition we find that · P(8, 3) = 8 ! / (8 - 3) ! = 8 ! / 5 ! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / (5 * 4 * 3 * 2 * 1) = 8 * 7 * 6. OK, P(8, 3) is the number of ordered choices. But what if, as in this case, we are making unordered choices? That is, what if the order in which the choices are made doesn’t matter? Any given set of 3 items could have been chosen in 3 ! = 3 * 2 * 1 = 6 different orders. · So the number of ordered choices of 3 items is 3 ! = 6 times as great as the number of unordered choices. · Thus the number of unordered choices is 1 / 6 as great at the number of ordered choices. · To get the number of unordered choices we therefore divide the number of ordered choices by 6. · Remember that we arrived at the number 6 from the fact that there are 3 ! = 6 ways to choose the same 3 items in different orders. Thus the number of unordered choices is · # of unordered choices = # of ordered choices / (# of ways a given set of chosen objects can be ordered). · # of unordered choices = P(8, 3) / 3 !. We call the number of unordered choices C(8, 3), the number of combinations of 3 objects chosen without replacement from 8. Therefore · C(8, 3) = P(8, 3) / 3 !. Since P(8, 3) = 8 ! / (8 - 3) !, we have · C(8, 3) = (8 ! / (8 - 3) ! ) / 3 !, which by the rule for dividing a fraction by a number simplifies to · C(8, 3) = 8 ! / [ (8 - 3)! * 3 ! ]. OK, in summary we divide 8 ! by [ (8 - 3) ! * 3 ! ] · Dividing 8 ! by (8 - 3) ! we are left with the first three factors 8 * 7 * 6, giving us the number of ordered choices. · When we then divide by 3 ! , which is the number of orders in which 3 given objects could have been chosen, we are left with the number of unordered choices. More generally, if we want to know the number of ordered choices possible when r objects are chosen in order, without replacement from a collection of n objects, the number is P(n, r) = n ! / (n - r)! If we want the number of unordered choices, then we have to divide this result by the r ! ways the r objects could be ordered, and we get C(n, r) = n ! / [ (n - r) ! * r ! ]. The reasoning behind these expressions is identical to the reasoning we used when developing the expression for choosing 3 objects out of 8. Note also that the reasoning summarized here has been developed throughout the first three qa's and the corresponding queries and sections of the text. A review of some or all of those sources might provide additional reinforcement for these ideas. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q011. In terms of factorials, how would we express the number of possible unordered collections of 5 objects chosen from 16? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 16! / [ ( 16 - 5 ) ! * 5 ! ] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 16 ! / ( 16 - 5) ! Possible ordered sets of 5 objects chosen from the 16. There are 5 ! ways to order any unordered collection of 5 objects. There are thus 16 ! / [ ( 16 - 5 ) ! * 5 ! ] possible unordered collections of 5 objects from the 16. STUDENT QUESTION: 16!/(16-5)!*5! because there are 5 ways to do this for UNORDERED collections, correct? INSTRUCTOR RESPONSE: Close, but to clarify the terminology: There are 5 ! different orders in which the same unordered collection could have been chosen. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q012. In terms of factorials, how would we express the number of possible unordered collections of r objects chosen from n objects? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: n ! / [ r ! * ( n - r ) ! ] ??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are P(n, r) = n ! / ( n - r ) ! possible ordered collections of r objects. There are r ! ways to order any unordered collection of r objects. There are thus P ( n, r ) / r! = n ! / [ r ! * ( n - r) ! ] possible unordered collections of r objects chosen from n objects. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q013. When we choose objects without replacement and without regard to order, we say that we are forming combinations as opposed to permutations, which occur when order matters. The expression we obtained in the preceding problem gives us a formula for combinations: C ( n , r ) = P ( n, r) / r! = n ! / [ r ! ( n - r) ! ] This is the number of possible combinations, or unordered collections, of r objects chosen from a set of n objects. Show how you would use the formula to find the number of unordered selections of 3 numbered balls out of a set of 15, where the selections are made without replacement. Show how you would use the formula to find the number of unordered selections of 3 number balls out of a set of 15, if all three selected balls have double digits, again selecting without replacement. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 455??? 20??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The formula C ( n , r ) = P ( n, r) / r! = n ! / [ r ! ( n - r) ! ] gives the number of ways to select r objects from n, without replacement. The first question asks how many ways there are to select 3 objects from a set of 15, without replacement. So for this question, r = 3 and n = 15. The formula therefore gives us the result C(15, 3) = 15 ! / ( 3 ! * (15 - 3) ! ) = 15 ! / (3! * 12 !) = 15 * 14 * 13 * 12 * 11 * ... * 1 / ( (3 * 2 * 1) * (12 * 11 * ... * 1) ) = (15 * 14 * 13) * (12 * 11 * ... * 1) / ( (3 * 2 * 1) * (12 * 11 * ... * 1) ). The factor (12 * 11 * ... * 1) in the numerator is divided by the same factor in the denominator, giving us 1, so our expression becomes (15 * 14 * 13) * 1 / (3 * 2 * 1) = 15/3 * 14/2 * 13 * 1/1 = 5 * 7 * 13 * 1 = 455. Note that it is not appropriate in this course to use a calculator to simplify this expression, with the exception of the final multiplication 5 * 7 * 13. You need to show the simplification without reference to a calculator. Simplification is straightforward, just matching up quantities in the numerator with the appropriate quantities in the denominator. The second question asks how many ways there are to select 3 balls having double digits from the set of 15 balls. There are only six balls, numbers 10, 11, 12, 13, 14 and 15, having double digits. So the selection of the 3 balls would be restricted to a set of only 6 balls, not 15. So the formula would apply with r = 3 and n = 6. The result would be C(6, 3) = 6 ! / ( 3 ! * ( 6 - 3) ! ) = 6 ! / (3 ! * 3 !) = 6 * 5 * 4 * 3 * 2 * 1 / ( (3 * 2 * 1) * (3 * 2 * 1) ) = (6 * 5 * 4) * (3 * 2 * 1) / ( (3 * 2 * 1) * (3 * 2 * 1) ) = 6 * 5 * 4 / (3 * 2 * 1) = 6 / 3 * 5 * 4 / 2 = 2 * 5 * 2 = 20. A calculator would be completely inappropriate in evaluating C(6, 3). This calculation involves only matching up the (3 * 2 * 1) in the numerator with the (3 * 2 * 1) in the denominator, then matching up the divisions 6 / 3 and 4 / 2 which have whole-number results, and finally performing a simple multiplication. Extra information (easy to understand now; very useful to have done so now when start dealing with probability in the next chapter): There are 455 combinations of three balls from among the 15. 20 of these combinations consist solely of double-digit balls. So we would say that the probability of obtaining three double-digit balls when randomly selecting from the 15, without replacement, is P(three double-digit balls) = 20 / 455 = 4 / 91 (approximately equal to .044 or 4.4%). Self-critique (if necessary) ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q014. In selecting three balls from 15, without replacement, how many ways are there to get each of the following? Three balls all with single digits. Three balls all of which contain the digit 1. Three odd-numbered balls. Additional question, optional now but to your benefit in the near future: You know how many possible selections there are of 3 balls out of the 15. You have calculated how many possible selections have three 3-digit numbers, and how many have three single-digit numbers. How can you calculate the number that include at least one two-digit number and at least one single-digit number? What is your result? (Optional but easy if you have answered previous questions: What is the probability that this will occur?) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q015. In selecting three balls from 15, without replacement: From a collection of three different single-digit balls, how many different numbers can be obtained by placing the three balls in different orders? How many numbers are possible from a collection of three balls each containing a double-digit number? How many different ordered selections of 3 balls can be made from the 15? How many different ordered selections of 3 single-digit balls are possible? What therefore is the probability of obtaining a three-digit number from a random selection of 3 of the 15 balls? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course Mth 152 Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: There are two dice. Call one the 'first die' and the other the 'second die'. (Note that ‘die’ is the singular of ‘dice’). It is possible for the first die to come up 3 and the second to come up 6. It is possible for the first die to come up 4 and the second to come up 5. It is possible for the first die to come up 5 and the second to come up 4. It is possible for the first die to come up 6 and the second to come up 3. These are the only possible ways to get a total of 9. Thus there are 4 ways. We can represent these 4 ways as ordered pairs: (3,6), (4, 5), (5, 4), (6, 3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. In how many ways can we choose a committee of three people from a set of five people? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A committee when first chosen is understood to consist of equal individuals. The committee is therefore unordered. In choosing a committee of three people from a set of five people we are forming a combination of 3 people from among 5 candidates. The number of such combinations is C ( 5, 3) = 5 ! / [ 3 ! ( 5 - 3) ! ] = 5 ! / [ 3 ! * 2 ! ] = 5 * 4 * 3 * 2 * 1 / [ ( 3 * 2 * 1 ) * ( 2 * 1) ] = 5 * 4 / ( 2 * 1) = 5 * 2 = 10. STUDENT COMMENT: I really need to get this formula down!!! When are we supposed to use each formula?? INSTRUCTOR RESPONSE Combinations when order doesn't matter; permutations when it does. There's more, and your book states all the conditions. There are 10 possible 3-member committees within a group of 5 individuals. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. In how many ways can we choose a president, a secretary and a treasurer from a group of 10 people? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 720??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This choice is ordered. The order of our choice determines who becomes president, who becomes secretary and who becomes treasurer. Since we are choosing three people from 10, and order matters, we are looking for the number of permutations of three objects chosen from 10. This number is P(10, 3) = 10! / (10-3)! = 10! / 7! = 10 * 9 * 8 = 720. STUDENT COMMENT: I am so lost on the formulas, I have no clue which one to use where. These last two questions looked just a like to me. INSTRUCTOR RESPONSE: Between these two problems, order matters on one, it doesn't matter on the other. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. In how many ways can we arrange six people in a line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 720??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 6 ! = 720 possible orders in which to arrange six people. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. In how many ways can we rearrange the letters in the word 'formed'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 720??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are six distinct letters in the word 'formed'. Thus we can rearrange the letters in 6 ! = 720 different ways. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. In how many ways can we rearrange the letters in the word 'activities'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5040??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 10 letters in the word 'activities', but some of them are repeated. There are two t's and three i's. If we think of these 10 letters as being placed on letter tiles, there are 10! ways to rearrange the tiles. However, not all of these 10 ! ways spell different words. For any spelling the three 'i' tiles can be arranged in 3 ! = 6 different ways, all of which spell same word. And for any spelling the two 't' tiles can be arranged in 2 ! = 2 different ways. We must thus divide the 10! ways to arrange the ten tiles by the 3! ways in which the three i tiles might be ordered, and then divide this result by the 2! ways in which the three t tiles might be ordered, leading to the conclusion that 10 ! / ( 3 ! * 2 !) different ‘words’ are possible.. STUDENT COMMENT: I didn’t think about this because no where did it say the tiles cannot be repeated. And it really doesn’t say the new arrangement needs to make a new word INSTRUCTOR RESPONSE: The example invokes 10 tiles. Any rearrangement of the 10 tiles forms a word, though most don't form a word in any known language. If the 10 tiles all contain different letters, then every different rearrangment of the tiles forms a different word, and the number of possible words is equal to the number of possible reorderings of the tiles. If two or more of the tiles contain the same letter, then a different ordering of the tiles doesn't necessarily form a different word. In this example there are three i's, on three different tiles. If you switch two of those tiles, then you have a different arrangement of the actual tiles but the word remains the same. So there are more arrangements of tiles than there are different words. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. In how many ways can we line up four people, chosen from a group of 10, for a photograph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5040??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We are arranging four people chosen from 10, in order. The number of possible arrangements is therefore P ( 10, 4) = 10! / ( 10-4)! = 10 ! / 6 ! = 10 * 9 * 8 * 7. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. In how many ways can we get a total greater than 3 when rolling two fair dice? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 33??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: It would not be difficult to determine the number of ways to get totals of 4, 5, 6, etc. However it is easier to see that there is only one way to get a total of 2, which is to get 1 on both dice; and that there are 2 ways to get a total of 3 (we can get 1 on the first die and 2 on the second, or vice versa). So there are 3 ways to get a total of 3 or less when rolling two dice. Since there are 6 possible outcomes for the first die and 6 possible outcomes for the second, there are 6 * 6 = 36 possible outcomes for the two dice. Of the 36 total possible outcomes, we just saw that 3 give a total of 3 or less, so there must be 36 - 3 = 33 ways to get more than 3. STUDENT QUESTION First you said there are two ways to get three (we can get 1 on the first die and 2 on the second, or vice versa). Then you say in the bullet there is three ways. Which one is it? INSTRUCTOR RESPONSE There are two ways to get a total of 3. There are three ways to get a total of 3 or less. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q009. A committee consists of 5 men and 7 women. In how many ways can a subcommittee of 4 be chosen if the number of men and women must be equal? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 210??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If the numbers of women and men are equal, then there must be 2 of each. Recall that a committee is regarded as unordered. If order doesn't matter there are C(5, 2) ways to choose 2 men out of 5, and C(7, 2) ways to choose 2 women out of 7. We have to choose 2 men AND we have to choose 2 women, so by the Fundamental Counting Principal there are C(5, 2) * C(7, 2) = 10 * 21 = 210 possible subcommittees. ********************************************* Question: `q010. In how many ways can we get a total of 10 or more when rolling two dice? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6???