Using Permutations and Combinations in Probability - Statistics and Probability

1. A raffle will award 1st, 2nd, and 3rd prizes to three different people from 9 entrants. Order matters because 1st/2nd/3rd are different prizes. There are $9P3$ equally likely outcomes. What is the probability that Jordan wins any prize (1st, 2nd, or 3rd)?
   Answer: $\dfrac{3\cdot 8P2}{9P3}$ This problem requires counting the ways to award three distinct prizes to nine entrants to find the probability that Jordan wins any prize. Since prizes are different (1st, 2nd, 3rd), order matters—assigning 1st to someone differs from 2nd. The total possible outcomes are the permutations of selecting and assigning 3 out of 9, which is $9P3 = 9 \times 8 \times 7$. The favorable outcomes assign Jordan to one of the three prizes (3 choices), then assign the remaining two prizes from the other 8, which is $3 \times 8P2$. Thus, the probability is $(3 \times 8P2) / 9P3$. A common mistake is ignoring the distinct prizes and using combinations; always ask, 'Would switching two winners' prizes create a new outcome?' Here, yes, because 1st and 2nd are unique, so use permutations.
2. A password is created by arranging 4 different letters chosen from A, B, C, D, E, F (no repeats). Order matters because different arrangements form different passwords. If one password is chosen uniformly at random from all such 4-letter passwords, what is the probability that it starts with A?
   Answer: $\dfrac{5\mathrm{P}3}{6\mathrm{P}4}$ This problem involves counting 4-letter passwords from 6 letters to find the probability it starts with A. Order matters, as different sequences are different passwords, so we use permutations. Total passwords: 6P4 = 6 × 5 × 4 × 3. Favorable ones start with A, then arrange 3 from the remaining 5: 5P3 = 5 × 4 × 3. The probability is \frac{5P3}{6P4}. A common error is using combinations, ignoring order. Ask: would switching two letters create a new password? Yes, confirming permutations.
3. Seven runners compete in a race. Gold, silver, and bronze medals are awarded to the top 3 finishers. Order matters because gold/silver/bronze are different. There are $7P3$ equally likely podium outcomes. What is the probability that Lina wins a medal (finishes in the top 3)?
   Answer: $\dfrac{3\cdot 6P2}{7P3}$ This problem requires counting the podium outcomes for seven runners awarding distinct medals to find the probability Lina gets a medal. Since gold, silver, bronze are different, order matters—first place differs from second. The total possible outcomes are the permutations of assigning top 3 out of 7, which is $7P3 = 7 \times 6 \times 5$. The favorable outcomes place Lina in one of the three spots (3 choices), then assign the other two from the remaining 6, which is $3 \times 6P2$. Thus, the probability is $(3 \times 6P2) / 7P3$. A common mistake is treating finishes as unordered sets; always ask, 'Would switching two runners' medals create a new outcome?' Here, yes, because gold and silver are distinct, so use permutations.
4. From 10 students (6 girls and 4 boys), a 3-student committee is chosen at random. Because it is a committee, order does not matter. What is the probability that the committee has exactly 2 girls and 1 boy?
   Answer: $\dfrac{\binom{6}{2}\binom{4}{1}}{\binom{10}{3}}$ This problem asks for the probability of selecting a committee with exactly 2 girls and 1 boy. Since it's a committee, order doesn't matter - the committee {Alice, Beth, Charlie} is the same as {Beth, Charlie, Alice}. The total possible outcomes are C(10,3) = 120 ways to choose any 3 students from 10. The favorable outcomes require choosing 2 girls from 6 AND 1 boy from 4, which gives C(6,2) × C(4,1) = 15 × 4 = 60 ways. Therefore, the probability is [C(6,2) × C(4,1)]/C(10,3) = 60/120 = 1/2. A common error is using permutations for committees - ask yourself: "Does rearranging the committee members create a new committee?" No, so use combinations.
5. Ten names are entered into a raffle. Three distinct prizes (1st prize, 2nd prize, 3rd prize) are awarded to three different people, and order matters because the prizes are different. What is the probability that Harper wins 1st prize and Quinn wins 2nd prize?
   Answer: $\dfrac{1}{90}$ This raffle problem involves awarding three distinct prizes (1st, 2nd, 3rd) to 10 people, where Harper must win 1st and Quinn must win 2nd. Since the prizes have different values, order matters—winning 1st prize is different from winning 2nd prize. The total ways to award three distinct prizes to 10 people is 10 × 9 × 8 = 720 (using permutations). For favorable outcomes where Harper wins 1st and Quinn wins 2nd, we have 1 choice for 1st prize (Harper), 1 choice for 2nd prize (Quinn), and 8 choices for 3rd prize (anyone except Harper and Quinn), giving us 1 × 1 × 8 = 8 ways. The probability is 8/720 = 1/90. A key insight: when positions have different values (like prize rankings), always use permutations—ask "Is Harper winning 1st the same as Harper winning 3rd?" No, so order matters!
6. A teacher randomly assigns 4 different students (A, B, C, D) to sit in 4 seats in a row. Order matters because different seatings are different outcomes. What is the probability that A and B sit next to each other (in either order)?
   Answer: $\dfrac{12}{24}$ This problem requires counting seating arrangements to find the probability that two specific students are adjacent. Since seats are in a row and order matters, we use permutations, as swapping positions creates a different seating. The total possible outcomes are the arrangements of 4 students, which is (4! = 24). The favorable outcomes treat A and B as a single block (in AB or BA order), leaving 3 units to arrange in ($3! \times 2 = 12$) ways. Thus, the probability is (\dfrac{12}{24}). A common mistake is ignoring that order matters and using combinations, undercounting distinct seatings. To decide, ask: 'Would switching A and B's positions create a new outcome?' If yes, like here where seats are distinct, use permutations.
7. From 9 books on a shelf, 4 different books are selected and arranged in a row on a desk. Because the arrangement in a row matters, order matters. What is the probability that two particular books (Book X and Book Y) are both included somewhere in the row?
   Answer: $\dfrac{2\mathrm{P}2\cdot 7\mathrm{P}2}{9\mathrm{P}4}$ This problem involves selecting and arranging books in a row, where order matters because different arrangements create different displays. The total possible outcomes are 9P4 = 9×8×7×6 = 3024 ways to arrange 4 books from 9. For favorable outcomes (both Book X and Book Y included), we first choose positions for X and Y (2P2 = 2 ways to arrange them in 2 of the 4 positions), then fill the remaining 2 positions with 2 of the other 7 books (7P2 = 7×6 = 42 ways). This gives 2P2 × 7P2 = 2 × 42 = 84 ways. Therefore, the probability is (2P2 × 7P2)/9P4 = 84/3024 = 1/36. Ask yourself: "Is X-Y-A-B different from Y-X-A-B?" Yes, so use permutations throughout.
8. A school selects a 4-student team from 9 students, where 5 students are experienced and 4 are new. Because it is a team, order does not matter. What is the probability that the team has at least 3 experienced students?
   Answer: $\dfrac{\binom{5}{3}\binom{4}{1}+\binom{5}{4}\binom{4}{0}}{\binom{9}{4}}$ This problem requires finding the probability of selecting a team with at least 3 experienced students. Since it's a team selection, order doesn't matter. The total possible outcomes are C(9,4) = 126 ways to choose 4 students from 9. For favorable outcomes, "at least 3 experienced" means exactly 3 experienced OR exactly 4 experienced students. This gives C(5,3)×C(4,1) + C(5,4)×C(4,0) = 10×4 + 5×1 = 40 + 5 = 45 ways. Therefore, the probability is [C(5,3)×C(4,1) + C(5,4)×C(4,0)]/C(9,4) = 45/126 = 5/14. When you see "at least," remember to add up all qualifying cases. Since rearranging team members doesn't create a new team, use combinations throughout.
9. A school selects 2 students from a group of 4 girls and 5 boys to represent the school at an event. Order does not matter. What is the probability that both selected students are girls?
   Answer: $\dfrac{\binom{4}{2}}{\binom{9}{2}}$ This problem requires counting student selections to find the probability both are girls. Since order does not matter in the selection, we use combinations, as the pair is what counts, not who is chosen first. The total possible outcomes are choosing any 2 from 9, which is (\binom{9}{2}). The favorable outcomes are choosing 2 girls from 4, which is (\binom{4}{2}). Thus, the probability is (\dfrac{\binom{4}{2}}{\binom{9}{2}}). A common mistake is using permutations for unordered selections, like treating the representatives as ordered. To check, ask: 'Would switching the two selected students create a new outcome?' If no, as in this group, use combinations.
10. A 4-digit code is formed by arranging 4 different digits chosen from 1 through 6, with no repetition. Order matters. What is the probability that the code starts with 1?
    Answer: $\dfrac{1}{6}$ This problem requires counting digit arrangements to find the probability the code starts with 1. Since it's a code and order matters, we use permutations, as different sequences are unique codes. The total possible outcomes are arranging 4 distinct digits from 6, which is $6 \times 5 \times 4 \times 3$. The favorable outcomes fix 1 first, then arrange 3 from the remaining 5, which is $1 \times 5 \times 4 \times 3$. Thus, the probability simplifies to $\dfrac{1}{6}$. A common mistake is using combinations for ordered codes, undercounting distinct arrangements. To avoid this, ask: 'Would switching two digits create a new code?' If yes, like here, use permutations.
11. A 4-digit code is formed by arranging 4 different digits chosen from $\{0,1,2,3,4,5,6\}$. Digits cannot repeat, and order matters because different arrangements make different codes. All such codes are equally likely. What is the probability that the code starts with 0?
    Answer: $\dfrac{6\mathrm{P}3}{7\mathrm{P}4}$ This problem involves creating 4-digit codes where digits cannot repeat and order matters. Since different arrangements create different codes (1234 ≠ 4321), we need permutations. The total possible outcomes are 7P4 = 7×6×5×4 = 840 different codes using 4 digits from 7 available. For favorable outcomes (codes starting with 0), we fix 0 in the first position and arrange the remaining 3 positions from the 6 remaining digits: 6P3 = 6×5×4 = 120 codes. Therefore, the probability is 6P3/7P4 = 120/840 = 1/7. Don't confuse this with combinations - ask: "Is 0123 different from 3210?" Yes, they're different codes, so order matters and we use permutations.
12. A jar contains 3 green candies and 5 yellow candies. Four candies are drawn without replacement, and order does not matter. What is the probability that all 3 green candies are drawn (and the 4th candy is yellow)?
    Answer: $\dfrac{\binom{3}{3}\binom{5}{1}}{\binom{8}{4}}$ This problem requires counting candy selections to find the probability of drawing all green plus one yellow. Since order does not matter, we use combinations, as the handful is a set, not a sequence. The total possible outcomes are choosing 4 from 8, which is (\binom{8}{4}). The favorable outcomes are all 3 green from 3 and 1 yellow from 5, given by ($\binom{3}{3} \times \binom{5}{1}$). Thus, the probability is (\dfrac{\binom{3}{3}\binom{5}{1}}{\binom{8}{4}}). A common mistake is using permutations for unordered draws, inflating counts with sequences. To check, ask: 'Would switching two candies in the draw create a new outcome?' If no, like this jar draw, use combinations.
13. A class has 12 students, including 5 who are left-handed. A group of 3 students is chosen at random to present. Since it is a group, order does not matter. What is the probability that at least 1 of the selected students is left-handed?
    Answer: $1-\dfrac{\binom{7}{3}}{\binom{12}{3}}$ This problem asks for the probability that at least 1 left-handed student is selected. Since it's a group selection, order doesn't matter. The total possible outcomes are C(12,3) = 220 ways to choose 3 students from 12. For "at least 1 left-handed," it's easier to use the complement: find the probability of NO left-handed students and subtract from 1. The number of ways to choose 3 students from the 7 right-handed students is C(7,3) = 35. So the probability of no left-handed students is C(7,3)/C(12,3) = 35/220 = 7/44. Therefore, the probability of at least 1 left-handed student is 1 - 7/44 = 37/44. When you see "at least one," consider using the complement approach: it's often simpler than counting multiple cases.
14. A drawer contains 5 black socks and 3 white socks. Two socks are drawn without replacement. Order does not matter because only the colors drawn matter. There are ${8\choose 2}$ equally likely 2-sock selections. What is the probability that the two socks are the same color?
    Answer: $\dfrac{{5\choose 2}+{3\choose 2}}{{8\choose 2}}$ This problem requires counting the ways to select two socks from eight (five black, three white) without replacement to find the probability they match in color. Since only colors matter, order does not matter—selecting black then white is the same as white then black. The total possible outcomes are the combinations of choosing 2 out of 8, which is C(8,2) = 28. The favorable outcomes are both black, C(5,2) = 10, or both white, C(3,2) = 3, totaling 13. Thus, the probability is [C(5,2) + C(3,2)] / C(8,2). A common mistake is using permutations when order is irrelevant, like counting ordered pairs; always ask, 'Would switching the order of socks create a new outcome?' Here, no, since we care only about the pair, so use combinations.
15. A bag contains 4 red marbles and 3 blue marbles. Three marbles are drawn without replacement. Order does not matter because only the set of marbles drawn is considered. What is the probability of drawing exactly 2 red marbles and 1 blue marble?
    Answer: $\dfrac{\binom{4}{2}\binom{3}{1}}{\binom{7}{3}}$ This problem requires counting sets of marbles drawn to find the probability of a specific color mix. Order does not matter, as only the combination of marbles counts, so we use combinations. The total ways to draw 3 from 7 is \binom{7}{3}. Favorable outcomes are exactly 2 red and 1 blue, counted as \binom{4}{2} \times \binom{3}{1}. The probability is \frac{\binom{4}{2} \binom{3}{1}}{\binom{7}{3}}. Commonly, people use permutations if they think of draw order, but that's incorrect here. To check, ask: would switching two drawn marbles create a new outcome? No, so combinations work best.
16. Six different photos are to be displayed in a row on a shelf. Order matters because different left-to-right orders are different displays. There are $6!$ equally likely arrangements. What is the probability that a particular photo (Photo X) is placed at one of the two ends of the row?
    Answer: $\dfrac{2\cdot 5!}{6!}$ This problem requires counting the arrangements of six distinct photos in a row to find the probability that Photo X is at one end. Since it's a left-to-right display, order matters—swapping positions changes the arrangement. The total possible outcomes are all permutations of the six photos, which is 6! = 720. The favorable outcomes place X at the left end (then arrange the other 5: 5!) or right end (another 5!), totaling 2 × 5! = 240. Thus, the probability is (2 × 5!) / 6!. A common mistake is undercounting by forgetting both ends; always ask, 'Would switching Photo X to the other end create a new outcome?' Here, yes, left and right are distinct, so include both cases separately.
17. Six runners (all different) compete in a race. The gold, silver, and bronze medals are awarded, so order matters. What is the probability that Alex wins gold and Blake wins silver (bronze can be any of the remaining runners)?
    Answer: $\dfrac{1}{6\cdot 5}$ This problem requires counting medal assignments to find the probability of specific winners for gold and silver. Since medals are distinct and order matters, we use permutations, as who gets which medal changes the outcome. The total possible outcomes are assigning 3 medals from 6 runners, which is ($6 \times 5 \times 4$). The favorable outcomes have Alex gold (1 way), Blake silver (1 way), and any of 4 for bronze. Thus, the probability simplifies to ($\dfrac{1}{6 \times 5}$). A common mistake is using combinations when orders are distinct, like treating medals as identical. To avoid confusion, ask: 'Would switching gold and silver create a new outcome?' If yes, as in this race, use permutations.
18. Five different books (A, B, C, D, E) are placed in a row in a random order. Order matters because different arrangements count as different outcomes. There are $5!$ equally likely arrangements. What is the probability that books A and B are adjacent (next to each other)?
    Answer: $\dfrac{2\cdot 4!}{5!}$ This problem requires counting the arrangements of five distinct books to find the probability that A and B are next to each other. Since the books are in a row, order matters—swapping two books creates a different arrangement. The total possible outcomes are all permutations of the five books, which is 5! = 120. The favorable outcomes treat A and B as a single block (which can be AB or BA, so 2 ways), then arrange this block with the other three books, giving 2 × 4! = 48 ways. Thus, the probability is (2 × 4!) / 5!. A common mistake is forgetting the 2 ways to order within the block, undercounting favorables; always ask, 'Would switching positions within the block create a new outcome?' Here, yes, AB differs from BA, so include both.
19. A committee of 3 is chosen from 7 people: 4 seniors and 3 juniors. Order does not matter because no roles are assigned. There are ${7\choose 3}$ equally likely committees. What is the probability that the committee has exactly 2 seniors and 1 junior?
    Answer: $\dfrac{{4\choose 2}{3\choose 1}}{{7\choose 3}}$ This problem requires counting the ways to choose a three-person committee from seven (four seniors, three juniors) to find the probability of exactly two seniors and one junior. Since no roles are assigned, order does not matter—the committee is just a group. The total possible outcomes are the combinations of choosing 3 out of 7, which is C(7,3) = 35. The favorable outcomes are choosing 2 seniors from 4 and 1 junior from 3, which is C(4,2) × C(3,1) = 6 × 3 = 18. Thus, the probability is [C(4,2) × C(3,1)] / C(7,3). A common mistake is using permutations for ordered selection when it's unnecessary; always ask, 'Would switching two members' positions create a new outcome?' Here, no, since there are no positions, so combinations work best.
20. A club has 8 members. Three distinct officer positions—President, Vice President, and Treasurer—will be filled by selecting 3 different members. Order matters because the positions are different. What is the probability that Alex is selected as one of the three officers?
    Answer: $\dfrac{3\cdot 7P2}{8P3}$ This problem requires counting the possible ways to assign three distinct officer positions from eight club members to find the probability that Alex is selected for one. Since the positions are different, order matters—assigning someone as President differs from Vice President. The total possible outcomes are the permutations of selecting and assigning 3 members out of 8, which is 8P3 = 8 × 7 × 6. The favorable outcomes are those where Alex is assigned to one of the three positions: there are 3 choices for Alex's position, and then 7P2 ways to assign the remaining two positions from the other 7 members. Thus, the probability is (3 × 7P2) / 8P3. A common mistake is using combinations instead, like treating selections as unordered, but this ignores the distinct roles; always ask, 'Would switching two people's positions create a new outcome?' Here, yes, because positions are unique, so permutations are appropriate.

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