#$&* course Mth 152 003. C(n,r) and P(n,r)
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Given Solution: When choosing 4 objects out of 12, there are 12 choices for the first, 11 choices for the second, 10 choices for the third and 9 choices for the fourth object. If the order matters there are therefore 12 * 11 * 10 * 9 possible outcomes. If the order doesn't matter, then we have to ask in how many different orders any given collection of 4 objects could be chosen. Given any 4 objects, there are 4 choices for the first, 3 choices for the second, 2 choices for the third and 1 choice for the fourth. There are thus 4 * 3 * 2 * 1 orders in which a given set of 4 objects could be chosen. We therefore have 12 * 11 * 10 * 9 / ( 4 * 3 * 2 * 1) possible outcomes when order doesn't matter. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q002. If order does not matter, then how many ways are there to choose 5 members of a team from 23 potential players? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (23 * 22 * 21 * 20 * 19) / (5 * 4 * 3 * 2 * 1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q003. In how many ways can we line up 5 different books on a shelf? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5 * 4 *3 * 2 * 1 = 120 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: It should be clear that there are 5 * 4 * 3 * 2 * 1 ways, since there are 5 choices for the first book, 4 for the second, etc.. If we multiply these numbers out we get 5 * 4 * 3 * 2 * 1 = 120. It might be a little bit surprising that there should be 120 ways to order only 5 objects. It’s probably even more surprising that if we double the number of objects to 10, there are over 3 million ways to order them (you should be able to verify this easily enough). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q004. The expression 5 * 4 * 3 * 2 * 1 is often written as 5 ! , read 'five factorial'. More generally if n stands for any number, then n ! stands for the number of ways in which n distinct objects could be lined up. Find 6 ! , 7 ! and 10 ! . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I didn't get the final answer on 10! because I was just going on mental math, but I believe I spelled it out clear enough. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q005. What do we get if we simplify the expression (10 ! / 6 !) ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 6 * 5 * 4 * 3 * 2 * 1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 10 ! / 6 ! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 6 * 5 * 4 * 3 * 2 * 1). We can simplify this by rewriting it as 10 * 9 * 8 * 7 * (6 * 5 * 4 * 3 * 2 * 1) / ( 6 * 5 * 4 * 3 * 2 * 1) = 10 * 9 * 8 * 7. We see that the 6 * 5 * 4 * 3 * 2 * 1 in the numerator matches the same expression in the denominator, so when divided these expressions give us 1 and we end up with just 10 * 9 * 8 * 7 * 1 = 10 * 9 * 8 * 7. Note that this is just the number of ways in which 4 objects can be chosen, in order, from a collection of 10 objects. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q006. We saw above that there are 23 * 22 * 21 * 20 * 19 ways to choose 5 individuals, in order, from 23 potential members. How could we express this number as a quotient of two factorials? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 23 * 22 * 21 * 20 * 19 = 23 ! / 18 ! confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q007. How could we express the number of ways to rank 20 individuals, in order, from among 100 candidates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 100 ! / (100 - 20) ! = 100 ! / 80 ! = 100 * 99 * 98 * … * 81 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 100 choices for the first candidate, 99 for the second, 98 for the third, etc.. For the 20th candidate there are 81 choices. You should convince yourself of this if you didn't see it originally. Our product is therefore 100 * 99 * 98 * ... * 81, which can be expressed as 100 ! / 80 !. Note that the denominator is 80 !, which can be written as (100 - 20)! . So the result for this problem can be written as 100 ! / (100 - 20) ! = 100 ! / 80 ! = 100 * 99 * 98 * … * 81. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q008. How could we express the number of ways to rank r individuals from a collection of n candidates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: n ! / ( n - r ) ! confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: P(8, 3) = 8 ! / ( 8 - 3) ! = 8 ! / 5 ! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 5 * 4 * 3 * 2 * 1) = 8 * 7 * 6 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: P(n, r) = n! / ( n - r) !. To calculate P(8, 3) we let n = 8 and r = 3. We get P(8, 3) = 8 ! / ( 8 - 3) ! = 8 ! / 5 ! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 5 * 4 * 3 * 2 * 1) = 8 * 7 * 6. This number represents the number of ways in which 3 objects can be chosen, in order, from 8 objects. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q010. In how many ways can an unordered collection of 3 objects be chosen from 8 candidates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8 * 7 * 6 / ( 3 * 2 * 1) = 4 * 7 * 2 = 56 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: P ( n, r ) / r! = 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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: g `q013. When we choose objects 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: g `q013. When we choose objects 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!