- Which person is fixed in the first position? - How many people remain for the other two positions? - List the possible orders of those people.

1. Lisa is fixed in the first position. 2. Tim and Max can fill the remaining two positions in \(2!=2\) orders. 3. The orders are Lisa–Tim–Max and Lisa–Max–Tim.

\(2\) orders: Lisa–Tim–Max and Lisa–Max–Tim

- Place the cards into the four digit positions one at a time. - How many cards are available before the first choice, and how does that number change? - Why does the fundamental counting principle multiply the successive numbers of choices?

a) There are \(4\) choices for the thousands digit, then \(3\) for the hundreds digit, \(2\) for the tens digit, and \(1\) for the ones digit. Thus, the total is \(4\cdot3\cdot2\cdot1=4!=24\). b) The fundamental counting principle multiplies the number of choices at successive stages, so the product \(4\cdot3\cdot2\cdot1\) counts every complete arrangement exactly once.

a) \(24\) numbers b) The successive numbers of choices are \(4,3,2,\) and \(1\), so \(4\cdot3\cdot2\cdot1=24\).

- Compare which office each person has in the two assignments. - Ask whether swapping the same two people changes the outcome being recorded.

1. The two assignments give the students different offices, so they are different outcomes. 2. Because the offices are distinct, switching the two selected people changes the assignment. Therefore, order matters when counting these assignments.

No. The assignments are different because the offices are different, so order matters.

- Does the name of the job matter, or only who is chosen? - Write the two jobs as column headings. How could you list the possible assignments below them?

1. The two jobs are different, so switching them changes who does which job. 2. The assignments are Leila choosing music and Jonah bringing snacks, or Jonah choosing music and Leila bringing snacks.

Yes. Either Leila chooses music and Jonah brings snacks, or Jonah chooses music and Leila brings snacks.

- How many permutations are there of \(n\) distinct objects? - Multiply the number of arrangements by the time for each arrangement. - Use \(3600\,\text{s}=1\,\text{h}\) to convert the final time.

a) The number of permutations of \(7\) distinct books is \(7!=7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=5040\). b) The total time is \(5040\cdot10\,\text{s}=50{,}400\,\text{s}\). Converting to hours gives \(50{,}400\div3600=14\).

a) \(5040\) arrangements b) \(14\) hours

- Which cards can be used in the hundreds place if the number must be less than \(600\)? - After one card is used, how many remain for the next position? - Multiply the numbers of choices for the three positions.

1. For the number to be less than \(600\), the hundreds digit must be \(2\) or \(4\), giving \(2\) choices. 2. After choosing the hundreds digit, \(3\) cards remain for the tens digit. 3. Then \(2\) cards remain for the ones digit. 4. By the fundamental counting principle, the total is \(2\cdot3\cdot2=12\).

\(12\) numbers

- Count the choices for each position. - For part b, track how the available choices decrease after each digit is used. - For part c, count the codes in which one digit is repeated four times.

a) Each of the four positions has \(6\) choices, so there are \(6^4=1296\) codes. b) For four different digits, the successive choices are \(6,5,4,3\), giving \(6\cdot5\cdot4\cdot3=360\) codes. c) There are \(6\) codes in which one digit is repeated four times. Therefore, \(P(\text{all digits equal})=\frac{6}{1296}=\frac{1}{216}\approx0.00463\).

a) \(1296\) b) \(360\) c) \(\frac{1}{216}\approx0.00463\)

- Use a factorial to arrange distinct people in a row. - For part b, count only the people whose positions are not fixed. - For part c, arrange the first three people and the remaining five people separately, then multiply.

a) Eight distinct people can be arranged in \(8!=40{,}320\) ways. b) A and B are fixed in the first two positions in that order, so the remaining six people can be arranged in \(6!=720\) ways. c) A, B, and C can be arranged in the first three positions in \(3!\) ways, and the remaining five people can be arranged in \(5!\) ways. Thus, \(3!\cdot5!=720\).

a) \(40{,}320\) b) \(720\) c) \(720\)

- Count all arrangements of \(8\) distinct objects. - Count the arrangements that are sorted by page count. - Identify the assumption made about the different arrangements.

1. The \(8\) distinct books can be arranged in \(8!=40{,}320\) equally likely orders. 2. Exactly \(2\) orders satisfy the condition: increasing page count and decreasing page count. 3. Therefore, \(P=\frac{2}{8!}=\frac{1}{20160}\approx 0.00004960\). 4. This is a theoretical probability calculation based on equally likely outcomes.

The probability is \(\frac{1}{20160}\approx 0.00004960\). It is a theoretical probability based on equally likely arrangements.

- Decide whether the selected students have distinct roles. - Compare an ordered assignment with an unordered group. - For part b, count one specified group among all equally likely groups.

a) In case (1), the roles are distinct, so there are \(15\cdot14\cdot13=2730\) assignments. In case (2), only group membership matters, so there are \(\binom{15}{3}=455\) groups. b) Exactly one of the \(455\) equally likely groups is the specified group. Therefore, \(P=\frac{1}{455}\).

a) Case (1): \(2730\); case (2): \(455\) b) \(\frac{1}{455}\approx0.22\%\)

- Does drawing every ticket give an arrangement of all the distinct tickets? - What changes in the number of free positions when the winning ticket has a fixed place? - Compare the number of allowed full orders with the number of all full orders.

1. All \(100\) tickets are different, so there are \(100!\) complete draw orders. 2. Fix the grand-prize ticket in tenth place. The remaining \(99\) distinct tickets can fill the other positions in \(99!\) orders. 3. Every complete order is equally likely. The probability is \(\frac{99!}{100!}=\frac{1}{100}\).

a) \(100!\) complete orders b) \(99!\) complete orders; the other \(99\) tickets fill the remaining positions. c) \(\frac{1}{100}=0.01\)

- First count all arrangements satisfying the required first position. - Then count the arrangements that also violate the “B not last” condition. - Subtract the violating arrangements before forming the probability.

1. With A fixed first, the remaining seven letters can be arranged in \(7!\) ways. 2. Among those arrangements, the ones with B last have A fixed first, B fixed last, and the remaining six letters arranged in \(6!\) ways. 3. Therefore, the favorable count is \(7!-6!=5040-720=4320\). 4. The probability is \(\frac{4320}{8!}=\frac{4320}{40320}=\frac{3}{28}\).

a) \(7!-6!=4320\) arrangements b) \(\frac{3}{28}\)

- Try counting orders of three different cards without any restrictions. - When making numbers rather than codes, which digits may be used first? - Which orders counted as codes must be excluded as numbers?

1. Every order of the three different digits is a code, giving \(3!=6\) codes. 2. A three-digit number cannot begin with \(0\). The hundreds place has \(2\) choices, followed by \(2\) orders for the remaining digits, giving \(2\cdot2!=4\) numbers. 3. The codes beginning with \(0\) are valid codes but do not represent three-digit numbers.

a) \(6\) codes b) \(4\) three-digit numbers c) A code can start with \(0\), but a three-digit number cannot.

- Keep M and N together while moving P to different positions. - Can M and N exchange places inside their pair?

1. The pair can appear as MN or NM. 2. Place the pair before or after P. The four rows are MNP, NMP, PMN, and PNM. 3. There are \(2\) placements of the pair and \(2\) orders within it, giving \(2\cdot2=4\) rows.

MNP, NMP, PMN, PNM; \(4\) rows. Treat the neighboring cards as one block, with two possible orders inside it.

- How many runners actually receive a spot? - After choosing the starter, how many choices remain for finisher? - What additional information would an order of all five runners record?

1. Only two of the five runners are assigned; the other three have no role. Using \(5!\) orders all five runners and overcounts the choices. 2. There are \(5\) choices for starter and then \(4\) remaining choices for finisher. 3. The number of assignments is \(5\cdot4=20\).

\(20\) assignments. \(5!\) incorrectly orders all five runners instead of filling only two spots.

- Follow one path from the start to an endpoint. What does that path describe? - Count endpoints under each first choice. - Which paths begin with Leila?

1. The tree has three branches for president, each followed by two branches for secretary. 2. Every endpoint represents one assignment, so there are \(3\cdot2=6\) assignments. 3. The assignments with Leila as president are Leila–Mateo and Leila–Zoe, in president–secretary order.

\(6\) assignments. With Leila as president: Leila–Mateo and Leila–Zoe.

- Compare the number of favorable arrangements with the total number of arrangements of the six distinct letters. - For part a, decide which positions are already determined before arranging the remaining letters. - For part b, think about how adjacency changes the number of independent positions while still allowing the two vowels to switch order.

a) There are \(6!=720\) total arrangements. Fix F in the first position and A in the second position; the remaining \(4\) letters can be arranged in \(4!=24\) ways. Therefore, \(P=\frac{24}{720}=\frac{1}{30}\). b) Regard A and O as one adjacent unit. That unit and the four consonants form \(5\) objects, giving \(5!=120\) arrangements. The vowels can occur as AO or OA, so there are \(2\cdot120=240\) favorable arrangements. Thus, \(P=\frac{240}{720}=\frac{1}{3}\).

a) \(\frac{1}{30}\) b) \(\frac{1}{3}\)

- First focus on the requirement that Ava and Ben occupy consecutive seats. - How could you count placements while keeping those two seats together as you arrange the other friends? - After fixing the adjacent pair as a unit, remember that the two people can exchange positions within it.

1. Regard Ava and Ben as one adjacent unit. Together with Clara, David, and Eli, there are \(4\) units to arrange. 2. The \(4\) units can be arranged in \(4!=24\) ways. 3. Ava and Ben can switch order within their adjacent unit, giving \(2\) internal orders. 4. Therefore, the number of seating arrangements is \(24\cdot2=48\).

\(48\) seating arrangements

- In part a, how many positions remain after the two specified books are fixed? - In part b, separate the two possible end positions for the math book. - For each possible end position, count the permutations of the other three books.

a) The math and history books are fixed at the two ends. The two remaining books can be arranged in \(2!=2\) ways. b) The math book can be at either end, giving \(2\) disjoint cases. In each case, the other three books can be arranged in \(3!=6\) ways. Therefore, the total is \(2\cdot3!=12\).

a) \(2\) arrangements b) \(12\) arrangements

- Count the total number of available characters. - Compare how the number of choices changes when repetition is allowed versus forbidden. - For part c, compare the number of all-digit codes with the total number of codes.

a) There are \(18\) available characters, and each of the five positions has \(18\) choices. Thus, \(18^5=1{,}889{,}568\). b) Without repetition, the successive numbers of choices are \(18,17,16,15,14\), so the number of codes is \(18\cdot17\cdot16\cdot15\cdot14=1{,}028{,}160\). c) There are \(10^5\) all-digit codes out of \(18^5\) total codes. Therefore, \(P(\text{all digits})=\frac{10^5}{18^5}=\left(\frac{5}{9}\right)^5=\frac{3125}{59049}\approx0.0529\).

a) \(1{,}889{,}568\) b) \(1{,}028{,}160\) c) \(\frac{3125}{59049}\approx 5.29\%\)

- Decide whether repetition is allowed in each part. - Decide whether order matters. - Use a product for ordered selections and a combination for an unordered group.

a) With repetition allowed and order recorded, each spin has \(20\) possible results, so there are \(20^3=8000\) sequences. b) With no repeated number, the successive choices are \(20,19,18\). Thus, there are \(20\cdot19\cdot18=6840\) sequences. c) When three distinct numbers are selected without regard to order, the number of groups is \(\binom{20}{3}=1140\).

a) \(8000\) b) \(6840\) c) \(1140\)

- Determine whether the selected people have distinct roles. - Determine whether order matters in each situation. - Determine whether a name may be repeated.

a) The three roles are distinct and no student can be repeated, so there are \(25\cdot24\cdot23=13{,}800\) assignments. b) The four representatives form an unordered group, so there are \(\binom{25}{4}=12{,}650\) groups. c) Each of the four positions has \(25\) choices and repetition is allowed, so there are \(25^4=390{,}625\) sequences.

a) \(13{,}800\) b) \(12{,}650\) c) \(390{,}625\)

- Ask whether an item can be selected more than once. - Ask whether changing the order creates a different outcome. - Identify the number of available choices and the number of selections.

a) \(n=10\), \(k=4\), repetition is allowed, and order matters. The number of codes is \(10^4=10{,}000\). b) \(n=10\), \(k=3\), repetition is not allowed, and order matters. The number of outcomes is \(10\cdot9\cdot8=720\). c) \(n=12\), \(k=2\), repetition is not allowed, and order does not matter. The number of groups is \(\binom{12}{2}=66\).

a) \(n=10\), \(k=4\), repetition allowed, order matters; \(10{,}000\) outcomes b) \(n=10\), \(k=3\), no repetition, order matters; \(720\) outcomes c) \(n=12\), \(k=2\), no repetition, order does not matter; \(66\) outcomes

- Compare the restrictions in parts a) and b): one forbids any reuse, while the other only restricts neighboring positions. - For part c, first decide what must be true of the two digits used by an alternating code. - After choosing the two digits, consider how the first position determines the rest of the pattern.

a) Without repetition, the four positions have \(6,5,4,3\) choices, so there are \(6\cdot5\cdot4\cdot3=360\) codes. b) The first position has \(6\) choices. Each later position has \(5\) choices because it cannot equal the immediately preceding digit. Thus, there are \(6\cdot5^3=750\) valid codes. c) There are \(\binom{6}{2}=15\) pairs of distinct digits. Each pair produces two alternating codes, depending on which digit comes first, so there are \(30\) favorable valid codes. Therefore, \(P=\frac{30}{750}=\frac{1}{25}=0.04\).

a) \(360\) b) \(750\) c) \(\frac{1}{25}=4\%\)

- For part a, count the choices for each square. - For part b, track how the number of available colors changes after each choice. - For part c, compare the desired event with the event counted in part b.

a) Each of the five positions has \(12\) choices, so there are \(12^5=248{,}832\) patterns. b) For five different colors, the successive choices are \(12,11,10,9,8\), giving \(12\cdot11\cdot10\cdot9\cdot8=95{,}040\) patterns. c) Use the complement that all five colors are different. Thus, \(P(\text{at least one repeat})=1-\frac{95040}{248832}=\frac{89}{144}\approx0.6181\).

a) \(248{,}832\) b) \(95{,}040\) c) \(\frac{89}{144}\approx 61.81\%\)

- Separate the decision of who is selected from the decision of which selected person gets which office. - For one fixed committee, how many orders can its members occupy in distinct offices? - Generalize the same two-stage counting process from \(4\) selected students to \(k\).

1. For \(n=9\) and \(k=4\), there are \(\binom{9}{4}=126\) possible committees. 2. Once one four-person committee is fixed, its four distinct members can be assigned to the four distinct offices in \(4!=24\) orders. 3. Therefore, the number of office assignments is \(126\cdot24=3024\), which is \(P(9,4)\). 4. In general, first choosing an unordered group of \(k\) people gives \(\binom{n}{k}\) choices. Each chosen group can then be ordered among the \(k\) distinct offices in \(k!\) ways. 5. Hence \(P(n,k)=\binom{n}{k}k!\).

a) \(126\) committees and \(3024\) office assignments b) Each fixed committee has \(4!=24\) assignments because its four members can be ordered among four distinct offices. c) Choosing the \(k\)-person set and then ordering those \(k\) people gives \(P(n,k)=\binom{n}{k}k!\).

- For each relay leg, ask how many runners remain available after earlier assignments. - The ordered choices continue for exactly \(k\) factors. - Compare the product with the beginning of the expansion of \(n!\).

1. For the first relay leg there are \(10\) runner choices, then \(9\) for the second leg, then \(8\) for the third. Thus there are \(10\cdot9\cdot8=720\) assignments. 2. In general, each assigned runner is no longer available. The successive numbers of choices are \(n,n-1,n-2,\ldots,n-k+1\), so the multiplication principle gives \(n(n-1)\cdots(n-k+1)\). 3. Expand \(n!\) as \(n(n-1)\cdots(n-k+1)(n-k)!\). Dividing by \((n-k)!\) removes the unused tail of the factorial, leaving the ordered-selection product. 4. If \(k=n\), then \(\frac{n!}{(n-n)!}=\frac{n!}{0!}=n!\).

a) \(720\) assignments b) \(n(n-1)\cdots(n-k+1)\) c) \(\frac{n!}{(n-k)!}\); when \(k=n\), the value is \(n!\).

- Start by counting the equally likely ordered flavor choices for all six teenagers. - For part a, track how many flavors remain available if every new choice must differ from earlier choices. - For part b, decide how many common flavors could produce the event.

a) There are \(10^6\) equally likely ordered sets of choices. For all different flavors, the successive numbers of choices are \(10,9,8,7,6,5\). Thus, \(P=\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5}{10^6}=0.1512\). b) There are \(10\) favorable ordered outcomes in which everyone chooses the same flavor, one for each flavor. Thus, \(P=\frac{10}{10^6}=0.00001\).

a) \(0.1512\) b) \(0.00001\)

- For part a, think about how the four specified songs can remain consecutive while the whole playlist is rearranged. - Account separately for the placement of that consecutive group and the order of the songs within it. - In part b, which positions are fixed by the pattern, and how many choices remain at each step?

a) Regard the \(4\) band songs as one consecutive unit. That unit and the other \(8\) songs form \(9\) objects, which can be arranged in \(9!\) ways. The \(4\) band songs can be arranged internally in \(4!\) ways. Thus, the number of play orders is \(9!\cdot4!=8{,}709{,}120\). b) Fill the first four positions in the stated order. There are \(4\) choices for the first band song, \(8\) and then \(7\) choices for the next two songs, and \(3\) choices for the second band song. Arrange the remaining \(8\) different songs in \(8!\) ways. The count is \(4\cdot8\cdot7\cdot3\cdot8!=27{,}095{,}040\).

a) \(8{,}709{,}120\) play orders b) \(27{,}095{,}040\) play orders

- List the pairs of positions that have exactly one position between them. - Account for the two possible orders of Anna and Ben. - For part b, count the left-end and right-end cases separately.

a) There are \(8\) position pairs with exactly one position between them: \((1,3)\) through \((8,10)\). Anna and Ben can switch places in \(2\) ways, and the other \(8\) people can be arranged in \(8!\) ways. Thus, the count is \(8\cdot2\cdot8!=645{,}120\). b) For both people in the first three positions, choose their two positions in \(\binom{3}{2}\) ways, arrange Anna and Ben in \(2!\) ways, and arrange the other \(8\) people in \(8!\) ways. The same count applies to the last three positions, and the two events are disjoint. Therefore, \(P=\frac{2\binom{3}{2}2!8!}{10!}=\frac{2}{15}\approx0.1333\).

a) \(645{,}120\) lineups b) \(\frac{2}{15}\approx0.1333\), or about \(13.33\%\)

- Determine whether the prizes make the order of winners important. - Use an unordered count for identical prizes and an ordered count for ranked prizes. - For one exact equally likely outcome, compare one favorable result with the full sample space.

a) In model (1), order does not matter because the prizes are identical, so there are \(\binom{20}{4}=4845\) winner groups. In model (2), the four places are distinct, so there are \(20\cdot19\cdot18\cdot17=116{,}280\) ordered outcomes. b) Each outcome within a model is equally likely. Therefore, the probabilities of one exact prediction are \(\frac{1}{4845}\approx0.000206\) for model (1) and \(\frac{1}{116280}\approx0.00000860\) for model (2).

a) Model (1): \(4845\); model (2): \(116{,}280\) b) Model (1): \(\frac{1}{4845}\approx0.0206\%\); model (2): \(\frac{1}{116280}\approx0.000860\%\)

- Count all length-\(n\) codes. - Count the codes that use every value exactly once. - Form a probability in terms of \(n\) and test small positive integers.

1. There are \(n^n\) possible codes. 2. A code with all different entries is a permutation of the \(n\) available values, so there are \(n!\) favorable codes. 3. Thus, \(P(\text{all different})=\frac{n!}{n^n}\). 4. Check positive integer values in order: \(n=1\) gives \(1\); \(n=2\) gives \(\frac{1}{2}\); \(n=3\) gives \(\frac{2}{9}\approx0.222\); \(n=4\) gives \(\frac{3}{32}=0.09375\). 5. The first value below \(0.10\) is \(n=4\).

\(n=4\)

- Use the complement that all birth months are different. - Reduce the number of available unused months for each additional student. - Test consecutive values of \(k\) until the probability crosses \(60\%\).

1. Use the complement that all \(k\) students have different birth months. 2. For \(k\le12\), \(P(\text{all different})=\frac{12}{12}\cdot\frac{11}{12}\cdots\frac{12-k+1}{12}\). 3. Check consecutive values near the threshold: For \(k=4\), \(P(\text{at least one shared month})=1-\frac{12\cdot11\cdot10\cdot9}{12^4}=\frac{41}{96}\approx 0.427\). For \(k=5\), \(P(\text{at least one shared month})=1-\frac{12\cdot11\cdot10\cdot9\cdot8}{12^5}=\frac{89}{144}\approx 0.618\). 4. The probability first exceeds \(0.60\) when \(k=5\).

\(k=5\)

- Consider the caps one at a time and track how many unused designs remain. - Use a complement for “at least two match.” - Think about why relative frequencies from finitely many trials vary.

a) The probability that all six designs are different is \(\frac{25}{25}\cdot\frac{24}{25}\cdot\frac{23}{25}\cdot\frac{22}{25}\cdot\frac{21}{25}\cdot\frac{20}{25}\approx0.52229\). b) The event “at least two match” is the complement, so its probability is \(1-0.52229\approx0.47771\). c) A finite simulation produces a relative frequency that varies randomly from one set of trials to another. With \(1000\) trials, the estimate should often be near the theoretical probability but usually will not equal it exactly.

a) Approximately \(0.52229\), or \(52.23\%\) b) Approximately \(0.47771\), or \(47.77\%\) c) Simulation estimates are subject to random variation.

- For all different results, reduce the number of available new outcomes after each spin. - Use a complement for at least one repeat. - Compare the number of spins with the number of possible results.

1. The probability that all \(4\) results are different is \(\frac{12}{12}\cdot\frac{11}{12}\cdot\frac{10}{12}\cdot\frac{9}{12}=\frac{11880}{20736}\approx 0.57292\). The probability of at least one repeat is the complement, approximately \(0.42708\). Therefore, all different is more likely. 2. There are only \(12\) possible results. By the pigeonhole principle, \(13\) spins must include a repeat, so the probability of all different results is \(0\).

1. All different is more likely: approximately \(0.57292\) versus \(0.42708\). 2. \(0\)

- Which features of this PIN assignment rule are needed for the collision calculation? - For all different PINs, track how many unused PINs remain at each assignment. - Use a complement for at least one match. - Think about how the number of possible matching pairs changes as employees are added.

a) If the system checks existing PINs and never assigns one twice, the assignments are no longer independent. The probability of a repeated PIN would then be zero for these \(50\) employees, so the birthday-problem model would not apply. b) The probability that all \(50\) PINs differ is \(\frac{10000\cdot9999\cdot9998\cdots9951}{10000^{50}}\approx0.88453\). c) Taking the complement gives \(P(\text{at least one match})=1-0.88453\approx0.11547\). d) With more employees, there are more pairs that could match, and each additional PIN must avoid a larger set of previously assigned PINs.

a) For example, the system could reject any PIN already assigned. This would remove independent PIN assignment and prevent collisions. b) Approximately \(0.88453\) c) Approximately \(0.11547\), or \(11.55\%\) d) More employees create more possible matching pairs.

- Represent each assignment as a permutation and compare each entry with its position. - For part b, can you count the event directly or through another event that is easier to list? - Consider what remains after three correct assignments.

1. The permutations with exactly two correct assignments are \((1,2,4,3)\), \((1,4,3,2)\), \((1,3,2,4)\), \((4,2,3,1)\), \((3,2,1,4)\), and \((2,1,3,4)\). The identity permutation \((1,2,3,4)\) has four correct assignments, so these seven permutations have at least two correct assignments. 2. For the complement event in part b, the no-correct permutations are \((2,1,4,3)\), \((2,3,4,1)\), \((2,4,1,3)\), \((3,1,4,2)\), \((3,4,1,2)\), \((3,4,2,1)\), \((4,1,2,3)\), \((4,3,1,2)\), and \((4,3,2,1)\). Thus there are \(9\) such permutations out of \(4!=24\). 3. Therefore, \(P(\text{at least one correct})=1-\frac{9}{24}=\frac{5}{8}=0.625\). 4. If three guests receive the correct drinks, the only remaining drink must belong to the only remaining guest, so all four assignments are correct.

a) \((1,2,4,3)\), \((1,4,3,2)\), \((1,3,2,4)\), \((4,2,3,1)\), \((3,2,1,4)\), \((2,1,3,4)\), \((1,2,3,4)\) b) \(\frac{5}{8}=0.625\) c) Three correct assignments force the fourth assignment to be correct.

- First count all ordered outcomes for three rolls. - Once three distinct values are selected, only one ordering is strictly increasing. - For exactly two different values, one value must repeat. How many positions can hold the other value?

1. There are \(12^3=1728\) ordered outcomes. For three different results, there are \(12\cdot11\cdot10=1320\) outcomes, so the probability is \(\frac{1320}{1728}=\frac{55}{72}\approx0.7639\). 2. Choose any \(3\) distinct values from \(12\). Each chosen set has exactly one strictly increasing order, so the probability is \(\frac{\binom{12}{3}}{1728}=\frac{220}{1728}=\frac{55}{432}\approx0.1273\). 3. Choose the value that occurs twice in \(12\) ways and the other value in \(11\) ways. Choose which of the \(3\) positions holds the single different value. The total is \(12\cdot11\cdot3=396\) ordered outcomes.

1. \(P(\text{all different})=\frac{55}{72}\approx0.7639\) 2. \(P(\text{strictly increasing})=\frac{55}{432}\approx0.1273\) 3. \(396\) ordered outcomes: choose the repeated value, the other value, and which of the three positions holds the other value.

- Count all ordered triples first. - For part 1, identify the number multisets that sum to \(5\) and count their distinct orders. - A strictly decreasing sequence uses three distinct numbers in one specific order. - For part 3, restrict every draw to the values \(1\) through \(4\).

There are \(10^3=1000\) equally likely ordered triples. 1. The triples with sum \(5\) are the three permutations of \((1,1,3)\) and the three permutations of \((1,2,2)\), for six favorable triples. Thus, \(P=\frac{6}{1000}=\frac{3}{500}\). 2. A strictly decreasing triple is determined by choosing three distinct numbers; each chosen set has exactly one decreasing order. Thus, \(P=\frac{\binom{10}{3}}{1000}=\frac{3}{25}\). 3. If the greatest number is at most \(4\), every draw must be one of \(1,2,3,4\). Thus, \(P=\frac{4^3}{10^3}=\frac{8}{125}\).

1. \(\frac{3}{500}=0.006\) 2. \(\frac{3}{25}=0.12\) 3. \(\frac{8}{125}=0.064\)

- An even integer must end in which possible digits? - Treat a final digit of \(0\) separately because the first digit cannot be zero. - In each case, update the number of available digits after every position is filled.

1. Split into two cases according to the last digit. 2. If the last digit is \(0\), the first digit has \(9\) choices, then the two middle digits have \(8\) and \(7\) choices. This gives \(9\cdot8\cdot7=504\). 3. If the last digit is one of \(2,4,6,8\), there are \(4\) choices for the last digit. The first digit then has \(8\) nonzero choices, and the two middle digits have \(8\) and \(7\) choices. This gives \(4\cdot8\cdot8\cdot7=1792\). 4. The cases are disjoint, so the total is \(504+1792=2296\).

\(2296\) integers

- The two offices are different roles, so reversing two students creates a different assignment. - Express the number of ordered assignments in terms of the unknown number of eligible students. - After solving the equation, keep only a feasible positive integer value.

1. There are \(n\) choices for president and then \(n-1\) choices for vice president, so \(n(n-1)=72\). 2. Rearranging gives \(n^2-n-72=0\). 3. Factor: \((n-9)(n+8)=0\). 4. A student count must be positive, so \(n=9\).

\(9\) students

- Count all ordered outcomes from four runs. - For part b, first find the probability that every generated value is different. - For part c, count the outcomes in which one value repeats on all four runs.

a) For \(n=6\), there are \(6^4\) ordered outcomes. The number with four different values is \(6\cdot5\cdot4\cdot3\). Thus, \(P=\frac{6\cdot5\cdot4\cdot3}{6^4}=\frac{5}{18}\approx0.2778\). b) The probability that all four numbers are different is \(\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\frac{n-3}{n}=\frac{(n-1)(n-2)(n-3)}{n^3}\). Taking the complement gives \(P(\text{at least two equal})=1-\frac{(n-1)(n-2)(n-3)}{n^3}\). c) There are \(n\) ordered outcomes in which all four numbers are the same, out of \(n^4\) total outcomes. Hence, \(\frac{1}{n^3}=0.008\), so \(n^3=125\) and \(n=5\).

a) \(\frac{5}{18}\approx0.2778\) b) \(1-\frac{(n-1)(n-2)(n-3)}{n^3}\) c) \(n=5\)

- For all different cards, track how many choices remain after each distinct result. - Use the complement of “all three are different.” - For the threshold, solve the inequality and check the neighboring integer values.

a) For \(n=10\), \(P(\text{all different})=\frac{10}{10}\cdot\frac{9}{10}\cdot\frac{8}{10}=0.72\). b) For general \(n\), \(P(\text{all different})=\frac{(n-1)(n-2)}{n^2}\). Therefore, \(P(\text{at least two identical})=1-\frac{(n-1)(n-2)}{n^2}=\frac{3n-2}{n^2}\). c) Solve \(\frac{(n-1)(n-2)}{n^2}>0.98\), which is equivalent to \(0.02n^2-3n+2>0\). The relevant boundary is approximately \(149.33\), so the least integer satisfying the inequality is \(n=150\). Checking the neighboring integers confirms that \(n=149\) is below \(98\%\) and \(n=150\) is above it.

a) \(0.72\) b) \(P(\text{at least two identical})=\frac{3n-2}{n^2}\) c) \(n=150\)

- Count the ordered sequences in which every roll gives a new result. - Use a complement for “at least one repeat.” - For the binomial claim, check whether the proposed success probability remains constant across trials.

1. There are \(8^5\) ordered sequences. For all results to be different, there are \(8\cdot7\cdot6\cdot5\cdot4\) valid sequences, so the probability is \(\frac{105}{512}\approx0.2051\). 2. The event in part b) is the complement of all results being different, so the probability is \(1-\frac{105}{512}=\frac{407}{512}\approx0.7949\). 3. Marek’s claim is false. If “success” means repeating any previous result, its probability changes depending on how many distinct values have already appeared. A binomial model with \(p=\frac{1}{8}\) would instead count occurrences of one specified number.

a) \(\frac{105}{512}\approx0.2051\) b) \(\frac{407}{512}\approx0.7949\) c) The claim is false because the probability of a repeat is not constant from roll to roll.