- First determine from the diagram how many rays are displayed. - List every pair of displayed rays systematically. - Do not count the same pair twice in reverse order.

1. The diagram shows four rays labeled \(a\), \(b\), \(c\), and \(d\). 2. Each smaller angle is determined by an unordered pair of rays. 3. The pairs are \((a, b)\), \((a, c)\), \((a, d)\), \((b, c)\), \((b, d)\), and \((c, d)\). 4. There are \(6\) different pairs, so there are \(6\) different smaller angles.

\(6\) angles

- Decide whether changing the order of the same selected projects would create a new group. - Use notation that represents choosing \(k\) objects from \(n\) objects without regard to order.

1. The order of the selected projects does not matter, so use a combination. 2. Choosing 4 projects from 9 is represented by \(\binom{9}{4}\).

\(\binom{9}{4}\)

- Compare the members in the two sets rather than the order in which their names are written. - Ask whether any person's role changes when the names are reordered.

1. No. Both descriptions contain exactly the same three people. 2. A combination does not treat a different ordering of the same selected objects as a new outcome.

No. They are the same committee, so order does not matter in a combination.

- Fix the number of tokens in jar A first, then list what can happen in jars B and C. - Remember that the jars are labeled, so \((2,1,0)\) and \((1,2,0)\) are different. - Check that every listed triple has nonnegative entries that add to \(3\).

1. Group the distributions by the number of tokens in jar A. 2. If \(a=3\): \((3,0,0)\). 3. If \(a=2\): \((2,1,0)\), \((2,0,1)\). 4. If \(a=1\): \((1,2,0)\), \((1,1,1)\), \((1,0,2)\). 5. If \(a=0\): \((0,3,0)\), \((0,2,1)\), \((0,1,2)\), \((0,0,3)\). 6. There are \(10\) distributions. The same count can be represented by arranging three stars and two separators.

\((3,0,0)\), \((2,1,0)\), \((2,0,1)\), \((1,2,0)\), \((1,1,1)\), \((1,0,2)\), \((0,3,0)\), \((0,2,1)\), \((0,1,2)\), \((0,0,3)\); \(10\) distributions

- Does the folder record which postcard was chosen first? - List pairs without writing the same two postcards in reverse order.

1. The choices are mountain–lake, mountain–desert, and lake–desert. 2. The folder does not record an order, so mountain then lake is the same choice as lake then mountain.

\(3\) choices: mountain–lake, mountain–desert, lake–desert. No, reversing two postcards does not create a new choice.

- First imagine Ari is already on the committee. Who could join Ari? - Then leave Ari out. Which pairs remain possible? - Could any committee be counted in both cases?

1. If Ari is included, choose one of the other three students. There are \(3\) committees. 2. If Ari is not included, choose two of the remaining three students. There are \(3\) committees. 3. Every committee belongs to exactly one of these cases, so the total is \(3+3=6\). This illustrates \(\binom{4}{2}=\binom{3}{1}+\binom{3}{2}\).

\(3\) include Ari; \(3\) do not; \(6\) committees total.

- First determine how many of the five digits must be \(2\). - Once that number is fixed, what choice of positions determines the entire number? - Use an unordered position choice rather than listing cases one at a time.

1. Five digits of \(1\) would have sum \(5\). Replacing a \(1\) with a \(2\) increases the sum by \(1\), so exactly two positions must contain \(2\), and the other three positions must contain \(1\). 2. Choose the two positions occupied by \(2\). This gives \(\binom{5}{2}=10\) possible numbers.

\(\binom{5}{2}=10\) numbers

- List the pairs systematically so none are repeated or omitted. - Does the order of the two students matter? - Compare your list with what the binomial coefficient counts.

a) The two-student sets are \(\{A,B\}\), \(\{A,C\}\), \(\{A,D\}\), \(\{B,C\}\), \(\{B,D\}\), and \(\{C,D\}\). There are \(6\) selections. b) \(\binom{4}{2}=\frac{4\cdot3}{2\cdot1}=6\), which matches the list because combinations count selections without regard to order. c) Selecting one student gives \(\binom{4}{1}=4\) choices.

a) \(\{A,B\}\), \(\{A,C\}\), \(\{A,D\}\), \(\{B,C\}\), \(\{B,D\}\), \(\{C,D\}\) b) \(\binom{4}{2}=6\), matching the six listed sets. c) \(\binom{4}{1}=4\)

- Does the order in which the four games are chosen affect the final set? - Are games selected with or without replacement? - Which counting model applies to an unordered selection?

1. The group selects \(4\) different games from \(14\), and order does not matter. 2. The number of selections is \(\binom{14}{4}=\frac{14\cdot 13\cdot 12\cdot 11}{4\cdot 3\cdot 2\cdot 1}=1001\).

There are \(1001\) possible sets of games.

- Does the order in which the athletes are selected matter? - Can an athlete be selected more than once? - Which formula counts subsets of a given size?

1. The team is an unordered selection of \(3\) people from \(10\). 2. The number of teams is \(\binom{10}{3}=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120\).

There are \(120\) possible teams.

- Identify which expressions form symmetry pairs. - Use \(\binom{n}{k}=\binom{n}{n-k}\) before doing unnecessary arithmetic. - Simplify factors before multiplying when a direct calculation remains.

a) \(\binom{18}{2}=\frac{18\cdot17}{2}=153\). b) By symmetry, \(\binom{18}{16}=\binom{18}{2}=153\). c) \(\binom{25}{3}=\frac{25\cdot24\cdot23}{3\cdot2\cdot1}=2300\). d) By symmetry, \(\binom{25}{22}=\binom{25}{3}=2300\).

a) \(153\) b) \(153\) c) \(2300\) d) \(2300\)

- Evaluate or simplify the binomial coefficients one expression at a time. - Compare the first two results after calculating them. - In part c, look for symmetry before computing either coefficient.

a) \(\binom{9}{2}=36\) and \(\binom{9}{3}=84\), so the sum is \(36+84=120\). b) \(\binom{10}{3}=120\). c) By symmetry, \(\binom{11}{9}=\binom{11}{2}\). Therefore, the quotient is \(1\).

a) \(120\) b) \(120\) c) \(1\)

- Does order matter within each selected group? - Which formula counts \(k\) objects selected from \(n\) without replacement? - For part b, identify the calculator function that evaluates a binomial coefficient.

a) \(\binom{9}{2}=\frac{9\cdot8}{2\cdot1}=36\). b) \(\binom{18}{7}=31{,}824\).

a) \(36\) b) \(31{,}824\)

- How many groups of \(11\) can be selected from \(15\)? - Does selection order matter? - Choosing \(11\) players is equivalent to choosing which \(4\) are left out. - How many of the possible groups match the predetermined group?

1. The number of possible eleven-player groups is \(\binom{15}{11}=\binom{15}{4}=1365\). 2. Exactly one of these equally likely groups is the predetermined group, so the probability is \(\frac{1}{1365}\approx 0.0007326\), or approximately \(0.073\%\).

\(\frac{1}{1365}\approx 0.073\%\)

- How many choices are available at each pulse position? - For a fixed number of positive pulses, think about which positions are positive. - “At least \(6\)” includes \(6\), \(7\), and \(8\).

a) Each of the \(8\) positions has \(2\) choices, so there are \(2^8=256\) sequences. b) Choose the \(3\) positions for positive pulses: \(\binom{8}{3}=56\). c) At least \(6\) positive pulses means \(6\), \(7\), or \(8\). The number of favorable sequences is \(\binom{8}{6}+\binom{8}{7}+\binom{8}{8}=37\). Thus, the probability is \(\frac{37}{256}\approx0.1445\).

a) \(256\) b) \(56\) c) \(\frac{37}{256}\approx0.1445\)

- Can symmetry make any coefficient easier to compute? - Use the factorial formula only as far as needed before canceling. - For part f, evaluate each binomial coefficient before comparing the two sides.

a) \(\binom{9}{3}=84\). b) By symmetry, \(\binom{15}{13}=\binom{15}{2}=105\). c) \(\binom{25}{2}=300\). d) By symmetry, \(\binom{50}{48}=\binom{50}{2}=1225\). e) By symmetry, \(\binom{2000}{1999}=\binom{2000}{1}=2000\). f) The left side is \(\binom{10}{3}+\binom{10}{4}=120+210=330\), and the right side is \(\binom{11}{4}=330\). The two sides are equal.

a) \(84\) b) \(105\) c) \(300\) d) \(1225\) e) \(2000\) f) Left side: \(330\); right side: \(330\).

- Identify the total number of fruits. - Translate each event into the number selected from each fruit type. - Check that the selected counts add to five.

\(A\): This event requires two apples, two pears, and one banana, so it matches expression (2). \(B\): This event requires all five fruits to come from the \(12\) apples and bananas, so it matches expression (1). \(C\): This event requires three apples and two bananas, so it matches expression (3).

\(A\) matches (2). \(B\) matches (1). \(C\) matches (3).

- Count all groups and the groups containing the specified student. - For the symmetry explanation, ask whether any student is favored before the random selection.

a) There are \(\binom{25}{4}\) possible groups. Groups containing Luke are formed by choosing the other \(3\) students from the remaining \(24\), so \(P=\frac{\binom{24}{3}}{\binom{25}{4}}=\frac{4}{25}=0.16\). b) By symmetry, each of the \(25\) students has the same chance of being selected into one of the \(4\) places, so each student’s selection probability is \(\frac{4}{25}\).

a) \(\frac{4}{25}=0.16\) b) Four of the \(25\) equally likely students are selected, so each student has probability \(\frac{4}{25}\).

- Count unordered selections with a binomial coefficient. - Calculate the number of possible selections for each format. - When exactly one selection wins, how does the total number of selections affect the probability?

1. Format A has \(\binom{42}{6}=5{,}245{,}786\) possible selections. 2. Format B has \(\binom{37}{7}=10{,}295{,}472\) possible selections. 3. Each format has exactly one winning selection. Therefore, the jackpot probabilities are \(\frac{1}{5245786}\) for Format A and \(\frac{1}{10295472}\) for Format B. Since Format A has fewer possible selections, it has the greater jackpot probability.

Format A: \(5{,}245{,}786\) selections Format B: \(10{,}295{,}472\) selections Format A has the greater jackpot probability.

- Try showing a three-trial outcome with a short pattern of symbols. - Does the sequence depend on the order you name its successful trials? - How many choices are you making from the ten trial positions?

1. One representation marks the success positions among \(10\) numbered positions. For the example, mark positions \(2\), \(5\), and \(9\), with the other positions left unmarked. An equivalent sequence is F-S-F-F-S-F-F-F-S-F. 2. Choosing the positions in a different order does not change the resulting pattern of successes and failures. 3. Choose the \(3\) success positions from the \(10\) trial positions: \(\binom{10}{3}=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120\).

a) For example, use ten positions, marking \(2\), \(5\), and \(9\) as successes. b) No. The same selected trial positions give the same outcome sequence in either order. c) \(\binom{10}{3}=120\) outcome sequences

- For each student, how many choices are there about volunteering? - How would you count groups with a fixed number of members? - Does the order of names change who belongs to a group?

1. Each of four members can volunteer or not, giving \(2^4=16\) possible groups. 2. Groups of size two are counted by \(\binom{4}{2}=6\). 3. A group is determined by its members, not by the order in which their names are listed.

a) \(16\). b) \(\binom{4}{2}=6\). c) The order of two members does not make a new group.

- A disjoint event can use only outcomes outside \(A\). - Count the available outcomes in the complement. - The event is an unordered subset.

1. The complement of \(A\) contains \(10-4=6\) outcomes. 2. A disjoint event \(B\) must choose all \(3\) outcomes from those \(6\). 3. The number is \(\binom{6}{3}=20\).

There are \(\binom{6}{3}=20\) such events.

- Think of the total \(7\) as identical units that must be divided among three variables. - How many separators are needed to create three groups? - Once stars and separators are combined, count which positions can hold the separators.

1. Represent the \(7\) identical units by seven stars. 2. Two separators divide the stars into the amounts assigned to \(x\), \(y\), and \(z\). 3. There are \(7+2=9\) positions containing seven stars and two separators. 4. Choose the two separator positions: \(\binom{9}{2}=36\).

\(36\) solutions

- Would listing the same three dancers in reverse order make a different team? - How many ways can the same three members be listed in order? - What should happen to the student's count once those repeats are recognized?

1. The product counts the same three dancers in every possible order, but an unordered team does not change when its members switch places. 2. Each team is counted \(3!=6\) times in the product. 3. Divide to correct the overcount: \(\frac{6\cdot5\cdot4}{3!}=20\) teams.

\(20\) teams. The student's count includes each three-person team \(3!=6\) times.

- What is the smallest number of tokens box A can have? - Can you turn the minimum requirement into a smaller counting problem? - How can you count distributions of identical objects among labeled boxes?

1. Put two tokens in box A first to meet the minimum. 2. Distribute the remaining three identical tokens freely among the three labeled boxes. 3. By stars and bars, the number is \(\binom{3+3-1}{3-1}=\binom{5}{2}=10\). 4. Removing two tokens from A turns each allowed distribution into one distribution of three tokens. Adding two back to A reverses this step, so every allowed distribution is counted once.

\(10\) distributions. Reserve two tokens for A and distribute the remaining three. Every allowed distribution corresponds to exactly one distribution of those three tokens: remove the two reserved tokens from A to get it, or add them back to reverse the step.

- Calculate the counts of committees as the committee size grows. - Are the numbers rising or falling before you reach size \(3\)? - What does choosing the members left out tell you about committees of sizes greater than \(3\)?

1. The counts for sizes \(0\), \(1\), \(2\), and \(3\) are \(1\), \(6\), \(15\), and \(20\). 2. They increase toward size \(3\); in particular, \(\frac{\binom{6}{3}}{\binom{6}{2}}=\frac{20}{15}=\frac{4}{3}>1\). 3. By \(\binom{6}{k}=\binom{6}{6-k}\), the counts for sizes \(4\), \(5\), and \(6\) are \(15\), \(6\), and \(1\). Thus the maximum is \(20\) for size \(3\).

\(1,6,15,20\). Symmetry completes the row as \(1,6,15,20,15,6,1\); the largest count is \(\binom{6}{3}=20\).

- Which two-student teams contain Hana? Which contain Mina? - Can one team appear on both lists? - How would you check the total against all possible two-student teams?

1. There are \(3\) two-student teams with Hana and \(3\) with Mina. 2. The team consisting of Hana and Mina appears in both counts. Subtract that one repeated team. 3. The correct total is \(3+3-1=5\). As a check, there are \(\binom{4}{2}=6\) teams in all, and only the Luis–Noah team has neither student.

\(5\) teams. The Hana–Mina team was counted twice, so one copy must be removed.

- What can the other two boxes hold if A is empty? - Does the same kind of count work when B is empty? - Can both conditions describe the same distribution?

1. If A is empty, the number of tokens in B can be any whole number from \(0\) to \(7\); C gets the rest. This gives \(8\) distributions. 2. If B is empty, there are also \(8\) distributions, determined by the number in A. 3. Both A and B are empty in exactly one distribution: all seven tokens are in C. That distribution was counted twice. 4. Subtract one repeated count: \(8+8-1=15\) distributions.

\(15\) distributions. The case with all seven tokens in C appears in both empty-box counts.

- First choose four students for one group. - Determine whether the remaining group requires another choice. - Account for the fact that the two groups have no labels.

1. Choose four of the eight students for one group: \(\binom{8}{4}=70\). 2. The remaining four students automatically form the other group. 3. Each grouping has been counted twice because choosing either group first produces the same pair of unlabeled groups. Therefore, \(70\div2=35\).

\(35\) groupings

- Does the order of the two selected students matter? - Once two students are selected, what is determined about the rest of the class? - Compare \(\binom{n}{k}\) with \(\binom{n}{n-k}\).

a) Since order does not matter, the number of pairs is \(\binom{15}{2}=\frac{15\cdot14}{2}=105\). b) The number of complementary groups of \(13\) students is \(\binom{15}{13}=105\). Every choice of \(2\) selected students uniquely determines the \(13\) not selected, illustrating \(\binom{n}{k}=\binom{n}{n-k}\). Thus, \(\binom{15}{2}=\binom{15}{13}\).

a) \(105\) b) \(105\). The counts are equal because \(\binom{15}{2}=\binom{15}{13}\); choosing the two selected students determines the thirteen not selected.

- How many balls remain when \(3\) are selected? - How are the selected and unselected sets related? - Compute the two binomial coefficients and compare them.

a) The number of three-ball selections is \(\binom{7}{3}=35\). b) The number of four-ball selections is \(\binom{7}{4}=35\). The claim is true because every set of \(3\) selected balls uniquely determines the complementary set of \(4\) balls left behind. Thus, \(\binom{7}{3}=\binom{7}{4}\).

a) \(35\) b) The claim is true because \(\binom{7}{3}=\binom{7}{4}=35\). Choosing \(3\) balls determines the \(4\) balls left behind.

- Count all selections of five tickets from \(30\). - For each event, decide what must be included or excluded from the selected set. - For part c, compare selections containing the specified ticket with all five-ticket selections.

a) There are \(\binom{30}{5}=142{,}506\) equally likely five-ticket selections. Exactly one selection contains all five winning tickets, so \(P=\frac{1}{142506}\). b) There are \(25\) nonwinning tickets. Thus, \(P=\frac{\binom{25}{5}}{\binom{30}{5}}=\frac{1265}{3393}\approx0.3728\). c) If the specified ticket is included, choose the other four tickets from the remaining \(29\). Thus, \(P=\frac{\binom{29}{4}}{\binom{30}{5}}=\frac{1}{6}\).

a) \(\frac{1}{142506}\) b) \(\frac{1265}{3393}\approx0.3728\) c) \(\frac{1}{6}\)

- Paths with a fixed number of successes can be counted by choosing the success positions. - A fixed beginning reduces the number of open positions. - How many successes have already occurred in \(F,S,F\)?

a) Choose the \(4\) success positions among \(10\): \(\binom{10}{4}=210\). b) In the fixed beginning \(F,S,F\), one success has already occurred. Seven positions remain, and exactly \(3\) more successes are needed. Thus, the number of qualifying paths is \(\binom{7}{3}=35\).

a) \(210\) b) \(35\)

- Think about the objects that are not selected. - How does a choice of included objects determine a choice of excluded objects? - For part c, compare two neighboring coefficients using their factorial forms. When is the next one larger?

a) Choosing the \(k\) included objects uniquely determines the \(n-k\) excluded objects, so the two counts are equal. b) Choosing \(38\) of \(40\) objects is equivalent to choosing the \(2\) objects left out. Thus, \(\binom{40}{38}=\binom{40}{2}\). c) The ratio of consecutive coefficients is \(\frac{\binom{12}{k+1}}{\binom{12}{k}}=\frac{12-k}{k+1}\). This is greater than \(1\) when \(k<6\) and less than \(1\) when \(k\ge6\). Thus the coefficients increase through \(k=6\) and then decrease. The unique maximum is \(\binom{12}{6}\).

a) Every choice of \(k\) included objects uniquely determines the \(n-k\) excluded objects. b) Choosing \(38\) objects to include is equivalent to choosing \(2\) objects to exclude. c) \(k=6\), because consecutive coefficients increase until that point and decrease afterward.

- Does the order in which members are chosen change the group? - In part b, count the choices from each subgroup separately. - For part c, identify what each object represents and whether it returns to the bag after selection.

a) Since order does not matter, the number of six-member groups is \(\binom{16}{6}=8008\). b) Choose \(3\) of the \(9\) women and \(3\) of the \(7\) men. The number of groups is \(\binom{9}{3}\binom{7}{3}=2940\). c) Use a bag with \(16\) distinct objects, one for each member: \(9\) of one type and \(7\) of another. Draw \(6\) objects without replacement and ignore the order of the draw.

a) \(8008\) groups b) \(2940\) groups c) Draw \(6\) distinct member-objects without replacement from a bag containing \(9\) objects of one type and \(7\) of another; order does not matter.

- Decide whether order matters in each selection. - For exactly \(2\) defective resistors, determine how many working resistors must also be selected. - “At least \(3\)” includes which separate cases? - For the bag model, represent each physical resistor by a distinct object.

1. Order does not matter, so the total number of four-resistor selections is \(\binom{25}{4}=12{,}650\). 2. For exactly \(2\) defective resistors, choose \(2\) of the \(5\) defective resistors and \(2\) of the \(20\) working resistors: \(\binom{5}{2}\binom{20}{2}=1900\). 3. “At least \(3\)” means exactly \(3\) or exactly \(4\) defective resistors, giving \(\binom{5}{3}\binom{20}{1}+\binom{5}{4}\binom{20}{0}=205\). 4. A matching bag model uses \(25\) distinct objects: \(5\) marked defective and \(20\) marked working. Draw \(4\) objects without replacement and ignore order.

a) \(12{,}650\) selections b) \(1900\) selections c) \(205\) selections d) Draw \(4\) objects without replacement and without regard to order from a bag containing \(5\) defective objects and \(20\) working objects.

- Count all five-person committees. - How many selected members must come from outside the teenage group? - How would you recognize two selections that name the same committee?

1. The total number of committees is \(\binom{25}{5}\). 2. A favorable committee contains three of the ten teenagers and two of the fifteen adults, so the number of favorable committees is \(\binom{10}{3}\binom{15}{2}\). 3. Therefore, \(P(\text{exactly three teenagers})=\frac{\binom{10}{3}\binom{15}{2}}{\binom{25}{5}}=\frac{60}{253}\approx 0.2372\).

\(\frac{60}{253}\approx 23.72\%\)

- Does order matter in a set of chosen side dishes? - In part b, focus on the dishes that are not selected. - For part c, consider the yes-or-no decision for each individual dish.

a) The number of selections of exactly \(4\) side dishes is \(\binom{12}{4}=495\). b) Choosing the \(8\) included dishes uniquely determines the \(4\) excluded dishes, so \(\binom{12}{8}=\binom{12}{4}\). c) Each of the \(12\) side dishes has two possibilities: chosen or not chosen. Thus, there are \(2^{12}=4096\) selections. Equivalently, \(\sum_{k=0}^{12}\binom{12}{k}=4096\).

a) \(495\) selections b) Choosing \(8\) dishes to include is equivalent to choosing \(4\) dishes to leave out, so \(\binom{12}{8}=\binom{12}{4}\). c) \(4096\) selections

- How many positions must be chosen for the correct answers? - “At least \(9\)” includes which two cases? - For the total, group answer patterns by the number of correct answers.

a) Choose the \(7\) question positions that are answered correctly: \(\binom{10}{7}=120\). A corresponding selection model uses \(10\) labeled question objects and selects \(7\) without replacement; the selected objects represent the correctly answered questions. b) “At least \(9\)” means exactly \(9\) or exactly \(10\) correct answers. Thus, the number of patterns is \(\binom{10}{9}+\binom{10}{10}=11\). c) Grouping all answer patterns by their number of correct answers gives \(\sum_{k=0}^{10}\binom{10}{k}=2^{10}=1024\).

a) \(120\) answer patterns; select \(7\) of \(10\) labeled question objects without replacement and without regard to order. b) \(11\) answer patterns c) \(\sum_{k=0}^{10}\binom{10}{k}=2^{10}=1024\)

- Choose members from the two categories separately. - For part b, compare the favorable team count with all six-person teams. - In part c, interpret each binomial coefficient before interpreting their sum.

a) Choose four of the fifteen college students and two of the ten working professionals: \(\binom{15}{4}\binom{10}{2}=61{,}425\). b) There are \(\binom{25}{6}=177{,}100\) total six-person teams. Therefore, \(P=\frac{61425}{177100}=\frac{351}{1012}\approx0.3468\). c) \(\binom{15}{6}\) counts teams made entirely of college students, and \(\binom{10}{6}\) counts teams made entirely of working professionals. Their sum counts teams consisting entirely of one group or the other.

a) \(61{,}425\) b) \(\frac{351}{1012}\approx34.68\%\) c) It counts six-person teams made entirely of college students or entirely of working professionals.

- Choose books from each genre separately. - “At least seven” includes exactly seven and exactly eight. - Add the favorable counts before comparing with all eight-book selections.

a) The factor \(\binom{18}{5}\) chooses five mystery novels, and \(\binom{12}{3}\) chooses three historical novels. Multiplying combines the two independent category choices into one eight-book selection. b) At least seven mystery novels means either seven mystery and one historical or eight mystery and no historical. The favorable count is \(\binom{18}{7}\binom{12}{1}+\binom{18}{8}\binom{12}{0}=425{,}646\). There are \(\binom{30}{8}=5{,}852{,}925\) total selections, so \(P=\frac{3638}{50025}\approx0.0727\).

a) The first factor chooses the mystery novels, the second chooses the historical novels, and the product combines the choices. b) \(\frac{3638}{50025}\approx7.27\%\)

- What information about the winners matters across the five drawings? - Would switching the order of two winners always create a different result? - Compare the chance of one possible winning order with the chance of all qualifying orders. - What is the probability that neither named participant wins a particular draw?

1. On each drawing, \(P(A)=\frac{1}{20}\), \(P(J)=\frac{1}{20}\), and the probability that someone else wins is \(P(O)=\frac{18}{20}\). 2. Choose the \(2\) drawings won by Alex and then the \(1\) drawing won by Jordan: \(\binom{5}{2}\binom{3}{1}=10\cdot 3=30\) arrangements. 3. Each qualifying arrangement has probability \(\left(\frac{1}{20}\right)^2\left(\frac{1}{20}\right)\left(\frac{18}{20}\right)^2\). 4. Therefore, the required probability is \(30\left(\frac{1}{20}\right)^3\left(\frac{18}{20}\right)^2=\frac{243}{80000}=0.0030375\).

\(\frac{243}{80000}=0.0030375\), or \(0.30375\%\)

- What must be true of the other draws if exactly two are red and two are blue? - Does the order of the six colors affect whether a result qualifies? - How could you count qualifying color sequences without listing them all? - Does replacement change any color's chance from one draw to the next?

1. The color probabilities on each draw are \(P(R)=0.2\), \(P(B)=0.3\), and \(P(Y)=0.5\). 2. If exactly \(2\) draws are red and exactly \(2\) are blue, the remaining \(2\) draws must be yellow. 3. Choose the \(2\) red positions and then the \(2\) blue positions: \(\binom{6}{2}\binom{4}{2}=15\cdot 6=90\) color sequences. 4. Each qualifying sequence has probability \((0.2)^2(0.3)^2(0.5)^2=0.0009\). 5. Therefore, the probability is \(90\cdot 0.0009=0.081\).

\(0.081\), or \(8.1\%\)

- Can you describe one team that meets both job requirements? - What would change if you replaced just one developer on that team? - Does naming the same five people in a different order create a new team?

1. Choose \(3\) of the \(12\) software developers: \(\binom{12}{3}=220\). 2. Choose \(2\) of the \(8\) marketing specialists: \(\binom{8}{2}=28\). 3. Each first choice can be paired with each second choice, so the multiplication principle gives \(220\cdot28=6160\) teams.

\(6160\) teams

- List the possible apple-and-pear counts that contain more apples than pears. - Count each possible case separately. - The cases do not overlap, so how should their counts be combined? - Selecting the fruits at the same time means order does not matter.

1. With \(4\) fruits total, more apples than pears means either \(3\) apples and \(1\) pear or \(4\) apples and \(0\) pears. 2. The number of selections with \(3\) apples and \(1\) pear is \(\binom{10}{3}\binom{6}{1}=120\cdot6=720\). 3. The number of selections with \(4\) apples and no pears is \(\binom{10}{4}\binom{6}{0}=210\cdot1=210\). 4. The cases are disjoint, so the total is \(720+210=930\).

\(930\) selections

- Find the probability of each of the three result categories on one roll. - Choose the positions of the two \(6\)s first. - After those positions are fixed, how many positions remain for the three \(1\)s? - Multiply the number of qualifying arrangements by the probability of one arrangement.

1. On each roll, \(P(6)=\frac{1}{6}\), \(P(1)=\frac{1}{6}\), and \(P(\text{other})=\frac{4}{6}=\frac{2}{3}\). 2. Choose the \(2\) positions for the \(6\)s and then the \(3\) positions for the \(1\)s: \(\binom{10}{2}\binom{8}{3}=45\cdot 56\) arrangements. 3. Each qualifying arrangement has probability \(\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^3\left(\frac{2}{3}\right)^5\). 4. Therefore, \(P=\binom{10}{2}\binom{8}{3}\left(\frac{1}{6}\right)^5\left(\frac{2}{3}\right)^5=\frac{280}{6561}\approx 0.0427\).

\(\frac{280}{6561}\approx 0.0427\), or about \(4.27\%\)

- Match each probability base with its exponent. - A binomial coefficient can choose positions for one category. - Why are only \(4\) positions available after the first choice?

1. The exponents show that event \(E\) contains exactly \(8\) light packages, \(3\) standard packages, and \(1\) heavy package. 2. The factor \(\binom{12}{8}\) chooses the \(8\) positions occupied by light packages among the \(12\) selections. 3. After those positions are chosen, \(4\) positions remain. The factor \(\binom{4}{3}\) chooses the \(3\) positions occupied by standard packages. The final position must contain the heavy package.

Event \(E\) is “exactly \(8\) light packages, \(3\) standard packages, and \(1\) heavy package.” The factor \(\binom{12}{8}\) chooses the light-package positions, and \(\binom{4}{3}\) chooses the standard-package positions from the remaining \(4\) positions.

- If two specified people are included, count the remaining open positions. - For a fixed composition, choose members from each category separately. - Interpret the subtraction in part b as removing one complementary case from all teams.

a) For event \(A\), Sophie and Leon are fixed members, so choose the other two team members from the remaining \(16\): \(P(A)=\frac{\binom{16}{2}}{\binom{18}{4}}\). For event \(B\), choose two of the ten girls and two of the eight boys: \(P(B)=\frac{\binom{10}{2}\binom{8}{2}}{\binom{18}{4}}\). b) The term \(\binom{8}{4}\) counts teams made only of boys. Subtracting these from all four-person teams counts teams containing at least one girl.

a) \(P(A)=\frac{\binom{16}{2}}{\binom{18}{4}}\); \(P(B)=\frac{\binom{10}{2}\binom{8}{2}}{\binom{18}{4}}\) b) The team contains at least one girl.

- For part a, think of a situation where four selected people occupy four distinguishable positions. - For part b, think of a selection where only the chosen set matters. - For part c, count favorable groups by choosing from the two categories separately.

1. For part a, one valid question is: “How many ways can four distinct officer roles be assigned to four of the \(25\) students?” The roles are distinct, so exchanging two selected students changes the assignment. This gives \(25\cdot24\cdot23\cdot22\). 2. For part b, one valid question is: “How many six-person groups can be selected from the \(25\) students?” Only membership matters, so the same six students are one group regardless of listing order. This gives \(\binom{25}{6}\). 3. For part c, choose three of the fifteen female students and two of the ten other students. Thus \(P(X=3)=\frac{\binom{15}{3}\binom{10}{2}}{\binom{25}{5}}=\frac{195}{506}\approx0.3854\).

a) One valid example: How many ways can four distinct officer roles be assigned to four of the \(25\) students? Order matters because changing which student has a role changes the assignment. b) One valid example: How many six-person groups can be selected from the \(25\) students? Order does not matter because only group membership matters. c) \(\frac{195}{506}\approx38.54\%\)

- First decide whether order matters for an unassigned committee. - In part b, compare one specified committee with all equally likely committees. - Distinct roles make different arrangements of the same three students count separately.

1. Without assigned roles, order does not matter, so the number of committees is \(\binom{20}{3}=1140\). 2. Exactly one of the \(1140\) equally likely committees contains the three class representatives. Thus, the probability is \(\frac{1}{1140}\approx0.000877\). 3. With three distinct roles, order matters. The number of assignments is \(20\cdot19\cdot18=6840\). Equivalently, \(\binom{20}{3}3!=6840\).

a) \(1140\) committees b) \(\frac{1}{1140}\approx0.000877\), or about \(0.0877\%\) c) \(6840\) role assignments

- Count all committees first. - Break “at least \(5\)” into the possible exact cases. - For part c, interpret the numerator before comparing it with the denominator.

a) There are \(\binom{25}{6}\) possible committees. For exactly \(3\) women and \(3\) men, \(P=\frac{\binom{15}{3}\binom{10}{3}}{\binom{25}{6}}\approx0.30830\). b) At least \(5\) men means either \(5\) men and \(1\) woman or \(6\) men. Thus, \(P=\frac{\binom{10}{5}\binom{15}{1}+\binom{10}{6}}{\binom{25}{6}}\approx0.02253\). c) The expression is the probability that all \(6\) selected committee members are women.

a) Approximately \(0.30830\), or \(30.83\%\) b) Approximately \(0.02253\), or \(2.25\%\) c) The probability that the committee consists entirely of women

- Combine the number-set choices with the separate color choice. - For part b, separate the player’s chosen numbers into matches and nonmatches. - For part c, count both nonmatching number sets and nonmatching colors.

a) There are \(\binom{30}{4}\) number sets and \(3\) colors, so the total number of results is \(3\binom{30}{4}=82{,}215\). b) To match exactly \(3\) numbers, choose \(3\) of the player’s \(4\) numbers and \(1\) of the other \(26\) numbers. The color must also match. Thus, \(P=\frac{\binom{4}{3}\binom{26}{1}}{3\binom{30}{4}}=\frac{104}{82215}\approx0.001265\). c) For no number matches, choose all \(4\) numbers from the other \(26\). There are \(2\) incorrect colors. Thus, \(P=\frac{2\binom{26}{4}}{3\binom{30}{4}}=\frac{29900}{82215}\approx0.36368\).

a) \(82{,}215\) b) Approximately \(0.001265\), or \(0.1265\%\) c) Approximately \(0.36368\), or \(36.37\%\)

- What makes one winning set different from another? - How many winning numbers must come from the player's set, and how many from outside it? - What does the simulation count tell you about the fraction of trials with exactly three matches?

1. There are \(\binom{49}{6}=13{,}983{,}816\) possible drawings. To obtain exactly \(3\) matches, choose \(3\) of the player's \(6\) numbers and \(3\) of the other \(43\) numbers. Thus, \(P(\text{exactly 3 matches})=\frac{\binom{6}{3}\binom{43}{3}}{\binom{49}{6}}=\frac{246820}{13983816}\approx 0.0176504\). 2. The relative frequency is \(\frac{17612}{1000000}=0.017612\). The absolute difference is \(\left|0.017612-0.0176504\right|\approx 0.0000384\), or about \(0.00384\) percentage points. 3. The law of large numbers states that, as the number of independent trials increases, the relative frequency tends to stabilize near the theoretical probability. The small difference after one million trials is consistent with this behavior.

1. \(P(\text{exactly 3 matches})\approx 0.0176504\), or about \(1.765\%\) 2. Relative frequency \(0.017612\); absolute difference approximately \(0.0000384\), or \(0.00384\) percentage points 3. With many trials, the relative frequency tends to approach the theoretical probability.

- Model each element by an include/exclude decision. - After fixing one required element, choose the remaining member of the subset. - Compare a subset's size with \(5\) minus that size.

a) Each of five elements is either included or excluded, giving \(2^5=32\) subsets. b) If \(1\) must be included and the subset has size \(2\), choose one of the other four elements. There are \(4\) favorable subsets, so the probability is \(\frac{4}{32}=\frac{1}{8}\). c) A five-element set cannot be split into a subset and complement of equal integer size, so the probability is \(0\).

a) \(32\) outcomes. b) \(\frac{1}{8}\). c) \(0\).

- Once you choose the Falcons, is there any further choice for the Owls? - Which Falcons team sizes are allowed? - Does the order of names within a team matter?

1. Choose the Falcons; the remaining students belong to the Owls. 2. Count all allowed Falcons team sizes: \(\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}\). 3. The number of splits is \(5+10+10+5=30\).

\(\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}=30\) splits.

- Condition on the set of possible head positions. - Count position pairs rather than full flip strings. - Identify which pairs are consecutive.

1. The head positions are uniformly one of \(\binom{4}{2}=6\) pairs. 2. Adjacent pairs are \(\{1,2\}\), \(\{2,3\}\), and \(\{3,4\}\), giving \(3\) favorable pairs. 3. The conditional probability is \(\frac{3}{6}=\frac{1}{2}\).

The probability is \(\frac{1}{2}\).

- A match is determined by two teams, and reversing their order does not create a new match. - Express the number of unordered pairs in terms of the unknown team count. - Keep only a positive integer solution that can represent a number of teams.

1. With \(n\) teams, each match is determined by choosing an unordered pair, so \(\binom{n}{2}=45\). 2. Then \(\frac{n(n-1)}{2}=45\), so \(n(n-1)=90\). 3. Rearranging gives \(n^2-n-90=0\), which factors as \((n-10)(n+9)=0\). 4. The feasible team count is \(n=10\).

\(10\) teams

- Choose the members of the teams one team at a time. - Track how many students remain after each choice. - Account for the fact that the three teams are unlabeled. - Determine how many orders produce the same three teams.

1. Choose three of the nine students for a first temporary team: \(\binom{9}{3}=84\). 2. Choose three of the remaining six students for a second temporary team: \(\binom{6}{3}=20\). 3. The remaining three students form the third team. This gives \(84\cdot20=1680\) ordered team selections. 4. Because the three teams are unlabeled, each grouping is counted \(3!=6\) times. Therefore, \(1680\div6=280\).

\(280\) groupings

- For parts a and b, think about which positions are occupied by the defective components. - Connect the count of success positions with the probability of any one success/failure pattern. - For part c, look for a way to place the good components so that you can see where a defective component could be inserted without creating adjacency.

a) Choose the \(2\) defective positions among \(10\): \(\binom{10}{2}=45\). b) The binomial probability is \(P(X=2)=\binom{10}{2}(0.05)^2(0.95)^8\approx0.0746\). c) Place the \(5\) good components first. They determine \(6\) possible separation positions around and between them. Choosing \(3\) of those positions for the defective components gives \(\binom{6}{3}=20\) sequences.

a) \(45\) b) \(P(X=2)\approx0.0746\) c) \(20\)

- Count all unordered pairs represented in the urn. - For part a, organize pairs by color. - For part b, list pairs systematically by their label sums. - For part c, count choices from the two required colors.

a) There are \(\binom{7}{2}=21\) equally likely pairs. Same-color pairs number \(\binom{3}{2}+\binom{2}{2}+\binom{2}{2}=5\), so \(P(\text{same color})=\frac{5}{21}\). b) The prime sums possible are \(2\), \(3\), \(5\), and \(7\). There is \(1\) pair with sum \(2\), \(4\) pairs with sum \(3\), \(4\) pairs with sum \(5\), and \(3\) pairs with sum \(7\). Thus there are \(12\) favorable pairs, so \(P(\text{prime sum})=\frac{12}{21}=\frac{4}{7}\). c) Choose one of the three red balls and one of the two green balls. There are \(3\cdot 2=6\) favorable pairs, so \(P(\text{one red and one green})=\frac{6}{21}=\frac{2}{7}\).

a) \(\frac{5}{21}\) b) \(\frac{4}{7}\) c) \(\frac{2}{7}\)

- Count all committees that remain possible after imposing the condition. - For the favorable committees, fix the named person first. - Use the same type of committee count in the numerator and denominator.

1. The number of committees containing at least one senior is \(\binom{12}{5}-\binom{9}{5}=792-126=666\). 2. Every committee containing Maya satisfies the condition, and there are \(\binom{11}{4}=330\) such committees. 3. The conditional probability is \(\frac{330}{666}=\frac{55}{111}\).

The probability is \(\frac{55}{111}\approx 0.495\).

- Under the condition, focus first on the possible pairs of face cards. - Separate favorable pairs by color. - Decide whether the selection of the remaining cards affects the ratio.

1. There are \(6\) red face cards and \(6\) black face cards. 2. The two face cards can be any pair from the \(12\), giving \(\binom{12}{2}=66\) possibilities. 3. Same-color pairs number \(2\cdot\binom{6}{2}=30\). 4. The three non-face cards are selected in the same number of ways for every face-card pair, so that factor cancels. 5. The conditional probability is \(\frac{30}{66}=\frac{5}{11}\).

The probability is \(\frac{5}{11}\approx 0.455\).

- Which digit cannot be zero in a three-digit number? - How might you count possible digit sums without listing each number? - Could one number belong to more than one zero-digit case? - What is different about counting only numbers with no zeros?

1. Write the hundreds digit as \(h\ge1\). Let \(a=h-1\); then \(a+t+u=8\), with all three values nonnegative. The sum prevents any digit from exceeding \(9\), so there are \(\binom{10}{2}=45\) possible numbers. 2. Counting zeros directly: with \(t=0\), there are \(9\) choices for \(a\) and \(u\). With \(u=0\), there are also \(9\). The number \(900\) belongs to both cases, so there are \(9+9-1=17\) numbers with at least one zero. 3. Checking by the opposite event: if all digits are positive, subtract \(1\) from each of the three digits. Their new sum is \(6\), giving \(\binom{8}{2}=28\) numbers with no zero. Hence \(45-28=17\) numbers have a zero. 4. Both counts agree, and the probability is \(\frac{17}{45}\).

There are \(45\) possible numbers and \(17\) with at least one zero, so the probability is \(\frac{17}{45}\). Counting the zero cases gives \(9+9-1=17\); counting the opposite event gives \(45-28=17\). Both methods count the same numbers with at least one zero.