#$&* course MTH 152 Running a bit behind. Shouldn't take long to finish the rest of query 3, qa 4, query 4, qa 5, query 5. 003. C(n,r) and P(n,r)
.............................................
Given Solution: When choosing 4 objects out of 12, there are 12 choices for the first, 11 choices for the second, 10 choices for the third and 9 choices for the fourth object. If the order matters there are therefore 12 * 11 * 10 * 9 possible outcomes. If the order doesn't matter, then we have to ask in how many different orders any given collection of 4 objects could be chosen. Given any 4 objects, there are 4 choices for the first, 3 choices for the second, 2 choices for the third and 1 choice for the fourth. There are thus 4 * 3 * 2 * 1 orders in which a given set of 4 objects could be chosen. We therefore have 12 * 11 * 10 * 9 / ( 4 * 3 * 2 * 1) possible outcomes when order doesn't matter. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok. Feeling good about this now. ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q002. If order does not matter, then how many ways are there to choose 5 members of a team from 23 potential players? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - 23 * 22 * 21 * 20 * 19 the product of this statement will be the number of outcomes if order was an issue. - Since it is not, we will consider how many members we’re considering (5) and will say that - 23 * 22 * 21 * 20 *19 / (5 * 4 * 3 * 2 * 1) is the number of outcomes, provided order isn’t an issue. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: ok ********************************************* Question: `q003. In how many ways can we line up 5 different books on a shelf? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - 5 * 4 * 3 * 2 * 1 ways. - I’m not entirely sure about this one. I feel a need to know how many books can be put on a shelf, and how many books to choose from. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Sure is. I found it wasn’t so hard to believe when you take x number of variables, let’s say letters of the alphabet. Consider the set {a, b, c, d, e} plugged into my system of checking the numbers in which the first variable, a, takes the place of each of the five slots. Then b takes place of each of the five slots; then c, and d, and so on. It really is no surprise there are as many different ways as there are; even though it does seem quite profound at times, it’s really interesting to see just how many options there are. Consider how many ways one would sort 500 cars in a parking garage, for example. (I won’t attempt to calculate this number, but if any students want to play around with how large these numbers can be, you can refer them to http://www.numberempire.com/factorialcalculator.php ) ------------------------------------------------ Self-critique Rating: Ok. See the link in my comment above. I generally don’t like to reference sites that end in .com, but in this case, I thought it was pretty cool.
.............................................
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): None. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. What do we get if we simplify the expression (10 ! / 6 !) ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - I first complicate the solution ( but inadvertently make it easier, isn’t that cool?) by extending the factorial values. - - 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 ------------------------------------------ 6 * 5 * 4 * 3 * 2 * 1 - - As you can see, [6!] lines up nicely with the numerator, and thus becomes eliminated. 10 * 9 * 8 * 7 = 90 * 56 = 5040. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): None ------------------------------------------------ Self-critique Rating: Ok ********************************************* 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: - I would theorize that the numerator represents the total number of the collection, and the denominator represents the objects chosen from that collection. - With that said, this could be written as (23!)/(5!) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Where does the 18 ! come from? I thought we were working with 5 objects taken out of a collection of 23? ** I note that, when writing it out on paper, that the relationship between 23! and 18! as a quotient relates to the number of choices made (5) from a group that totals (23) in number, and the remaining choices from the 5 chosen are 23 * 22 * 21 * 20 * 19 representing the total number of options for the first, second, third, fourth, and fifth choices. ------------------------------------------------ Self-critique Rating: OK. See comment above.
.............................................
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): Could one then theorize that, instead of (Total number of objects in collection) ----------------------------------------------------------------------------- (total number of objects being taken from the collection) This would become (Total number of objects in collection) ---------------------------------------------------------------------------------------------------- (Factorial of the difference of objects being taken from the total collection So that way it would explain why 80! is used instead of 21. Because 100-20 = 80…. Just saw where you mentioned the exact same thing in the Given Solution. Oh well, at least I got my thoughts down and clarified! Note: I’ve revised my notes to say the following: “Numerator represents the total number of a given collection. Denominator represents what is the remainder of the total after the given number of objects are chosen.” ------------------------------------------------ Self-critique Rating: Ok. See comment above. ********************************************* Question: `q008. How could we express the number of ways to rank r individuals from a collection of n candidates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Taking into consideration the theory above, I would surmise that - - (n)/(n-r) Where n is the total number and n-r is the difference\factorial. *** Note the use of the backslash in place of the forward-slash. *** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): My first answer showed everything right except for the factorial symbol. ------------------------------------------------ Self-critique Rating: ok ********************************************* 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: - Taking the form P(n,r) = n! / (n - r) into consideration. - P(8,3) then takes the form = 8! / (8 - 3)! - Stretched out, this number becomes - 8*7*6*5*4*3*2*1 ----------------------- 5*4*3*2*1 Or simply 8*7*6, or 336. Meaning there are 336 permutations of 3 objects chosen from a collection of 8 possible objects. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: P(n, r) = n! / ( n - r) !. To calculate P(8, 3) we let n = 8 and r = 3. We get P(8, 3) = 8 ! / ( 8 - 3) ! = 8 ! / 5 ! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 5 * 4 * 3 * 2 * 1) = 8 * 7 * 6. This number represents the number of ways in which 3 objects can be chosen, in order, from 8 objects. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Nailed it on the first try. ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q010. In how many ways can an unordered collection of 3 objects be chosen from 8 candidates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - I take note of what you mentioned previously: The expression n ! / ( n - r ) ! denotes the number of ways in which r objects can be chosen, IN ORDER, from among n objects. - If the collection is unordered, I wonder how this would change? - 8*7*6*5*4*3*2*1 / 5*4*3*2*1 = 8*7*6, or 336 permutations in an ORDERED collection. Since this is unordered, I will say that 336 / 3 * 2 * 1 = 336/6 or 56 unordered permutations. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Two in a row that were nailed on the first try! ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q010. How could the result of the preceding problem be expressed purely in terms of factorials? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - I note that the previous solution was 8 * 7 * 6 / ( 3 * 2 * 1) = 4 * 7 * 2 = 56. - ***This was taken from your Given Solution, although mine matched in value I selected this as a quick way to backtrack a bit *** - 8*7*6 is the shortened form of 8! Or 8 * 7 * 6 * 5 * 4 * 3 * 2 *1, but since this was eliminated by 5 * 4 * 3 * 2 * 1 (where 8-3 = 5, then 5!) we can call 8 * 7 * 6 the collection of 3 objects taken from 8 possible objects. So we say that, when given the unordered choices of 3 objects taken from a collection of 8, - C(8,3) = P(8,3) / 3! (when in fact we’re multiplying by * 3! here?) I seem to be drawing a connection by your example of 8*7*6 = 8! / 5! From P(8,3) Taking (8-3) = 5 then 8! / 5! Is this a logical comparison? Or is this dangerous? Please provide any critical corrections here. I often look for patterns and so forth to make connections and better understand concepts.
.............................................
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): See my comment after the dotted line (or asterisk line) ------------------------------------------------ Self-critique Rating: ok ********************************************* 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: - Going by the preceding example, I would first compare - 5 objects chosen from 16 And - 3 objects chosen from 8 To attempt to reinforce my comprehension. Going from my last example, I simply subtracted 8-3 = 5 to form 8! / 5!. Now, it follows from this particular comparison that 16 - 5 = 11 to form 16 ! / 11 !. Note that this is before reading ahead to the Given Solution to check my answer to see if my method holds any credibility. ******************************* I’m close, I believe. When making this comparison, I seem to have left out the following concept: C(16,5) = 16! / [ (16-5) ! * 5!] Now, how does one express this kind of factorial as a number? Doesn’t this somewhat coincide with the concept of unordered collections being 1/x times as numerous and ordered collections being *x as numerous?
.............................................
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): I’m getting closer and closer to having this down pat. See my comment in Your Solution. ------------------------------------------------ Self-critique Rating: ok ********************************************* 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: - Since the number of ordered collections is: - P(n,r) = n! / (n - r) ! I will state that the number of UNordered collections is: - - C(n,r) = n! / [(n - r) ! * r!] This coincides with the relationship between the number of ordered ways being ‘r’ times as numerous, and the inverse is true (the number of unordered collections is thus 1/r as numerous) and since this is dealing with a fraction, the multiplication is necessary to show the denominator being larger (and thus a smaller fraction) so it follows that, for example, 1/45 is smaller than 1/5, because of this inverse proportion regarding the denominator. The larger the denominator, the smaller the fraction. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I believe that my answer C(n,r) = n! / [(n - r) ! * r!] is essentially the same except for - your usage of P (n, r) and my usage of C(n, r) [to specify an unordered collection] - your usage of n ! / [ r ! * ( n - r) ! ] and my usage of n! / [(n - r)! * r!] - - The brackets then have the same values, just a different order, correct? - (a * b) = (b * a) Also, is P(n, r) = n! / (n - r) ! Equivalent to P(n, r) / r! (?) ------------------------------------------------ Self-critique Rating: Ok. See my comment above when comparing my answer to your Given Solution. ********************************************* 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: - If - C(n, r) = P (n, r)/ r! = n! / [(n - r) ! * r!] to show a possible number of unordered collections from r objects chosen from n objects - I will consider, at this point, that I will be selecting 3 balls from a possible 15, so : C (n, r) = n! / [(n - r)! * r!] Then C (15, 3) = 15! / [ (15 - 3)! * 3!] This would tell you that, for a selected 3 balls chosen from a possible 15 15! / [ (15 - 3)! * 3!] identifies the number of unordered combinations **In what way is the nature of the balls having 2-digits relevant?** I note that one needs to simplify these numbers to 15*14*13*…*1 / (3*2*1) * (12*11*10…*1) From which the (12*11*10…*1) is taken from both the numerator and denominator, leaving 15*14*13 / 3 * 2 *1 which nicely divides into 15/3 = 5, 14/2 = 7, and 13/1 = 13. So now we take the product of 5 * 7 * 13, which I simplify to 35 * 13, = 455 possible unordered combinations. ************ However the question also specified listing the number of unordered combinations if any of the 3 balls had 2 digits, so out of the set {1, 2, 3….15} only the numbers {10, 11, 12, 13, 14, 15} would be considered with the aforementioned restriction. 3 balls are chosen out of the possible 6, so C (n, r) = n! / [ (n - r) * r! ] C (6, 3) = 6 * 5 * 4 * 3 *2 * 1 / [ (6 - 3)! * 3!] Reduces to C ( 6, 3) = 6 * 5 *4 *3 *2 *1 / (3 * 2 * 1) * ( 3 * 2 * 1) And with chopping involved, C (6, 3) = 6 * 5 * 4 / (3 * 2 * 1) Which will form C (6,3) 6/3 = 2, 4/2 = 2, and 5/1 = 5. So, C (6,3) = 2 * 2 * 5 = 20 possible unordered combinations with 2-digit balls only. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Nailed it. ------------------------------------------------ Self-critique Rating: Ok ********************************************* 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: