Combinations — Practice Set
Bank code: STA-CMB · Section: statistics · 30 questions — Easy 5 · Medium 8 · Hard 12 · Extreme 5
A combination is a selection where order does not matter: C(n,r) = n!/[r!(n-r)!], C(n,r) = C(n,n-r), and every pairing problem is C(n,2) = n(n-1)/2.
Every question below is live in the question bank under the ID shown — the sheet and the portal are the same questions. Attempt a level with the clock running, then check Part C.
Part A — Questions
Quantitative Comparison — the four choices are always the same, so they are not reprinted: (A) Column A is greater · (B) Column B is greater · (C) The two quantities are equal · (D) The relationship cannot be determined from the information given
Level 1 · Easy — 5 questions · ~4 min
warm-up — these must be automatic
Q1 · STA-CMB-006 · MCQ · 45s
A pizza shop offers 8 different toppings. A customer chooses exactly 2 different toppings for a single pizza. Both toppings go on the same pizza, so the order in which they are chosen does not matter, and no topping may be repeated. How many different 2-topping pizzas are possible?
(A) 16
(B) 28
(C) 36
(D) 56
(E) 64
Q2 · STA-CMB-009 · MCQ · 40s
A set has 4 distinct elements. How many subsets does it have in total, counting the empty set and the set itself?
(A) 4
(B) 8
(C) 15
(D) 16
(E) 24
Q3 · STA-CMB-011 · QC · 50s
There are 5 distinct books. In each column no book may be used more than once.
Column A: The number of ways to choose 3 of the books to donate, where the order of choice does not matter Column B: The number of ways to place 3 of the books into display slots labelled 1, 2 and 3, where the slot each book occupies does matter
Q4 · STA-CMB-013 · Numeric Entry · 55s
A coach must select 5 players from a roster of 12 distinct players to start a game. All starters play the same position and hold no distinct roles, so the order of selection does not matter, and no player may be selected twice. How many different starting lineups are possible? Enter your answer as a number.
Numeric entry — write the number.
Q5 · STA-CMB-015 · Select all that apply · 60s
Which of the following expressions are equal to C(8, 3)? Select all that apply.
(A) C(8, 5)
(B) 8! / (3! x 5!)
(C) (8 x 7 x 6) / 6
(D) 8 x 7 x 6
(E) C(7, 2) + C(7, 3)
Level 2 · Medium — 8 questions · ~10 min
two or three steps, one planted trap each
Q6 · STA-CMB-016 · MCQ · 70s
A committee of 4 must be chosen from 6 men and 4 women, all 10 people distinct. The committee members hold no distinct roles, so order does not matter, and no person may serve twice. How many committees contain exactly 2 men and exactly 2 women?
(A) 21
(B) 60
(C) 90
(D) 180
(E) 210
Q7 · STA-CMB-019 · MCQ · 85s
A test has 5 questions in Section A and 4 questions in Section B; all 9 questions are distinct. A student must answer exactly 4 of the 9 questions, and at least 2 of them must come from Section A. Only the set of questions answered matters, not the order in which they are answered. How many different selections are possible?
(A) 60
(B) 100
(C) 105
(D) 125
(E) 126
Q8 · STA-CMB-029 · MCQ · 85s
A club has 12 distinct members. It must form a finance committee of 3 members and a social committee of 4 members. No member may serve on both committees, and within each committee all members hold identical roles, so order does not matter. In how many ways can the two committees be formed?
(A) 220
(B) 495
(C) 27720
(D) 108900
(E) 3991680
Q9 · STA-CMB-024 · QC · 75s
n is an integer greater than 2.
Column A: C(n, 2) Column B: C(n-1, 2) + (n - 1)
Q10 · STA-CMB-034 · QC · 65s
A 10-person committee must pick a 3-person subcommittee. All 10 people are distinct, and in both columns order does not matter.
Column A: The number of ways to choose the 3 people who WILL serve on the subcommittee Column B: The number of ways to choose the 7 people who will NOT serve on the subcommittee
Q11 · STA-CMB-018 · Numeric Entry · 65s
A 4-person delegation is to be selected from 9 distinct candidates. Two particular candidates, Priya and Raj, must both be included. The delegates hold no distinct roles, so order does not matter, and no candidate may be selected twice. In how many ways can the delegation be selected? Enter your answer as a number.
Numeric entry — write the number.
Q12 · STA-CMB-030 · Numeric Entry · 70s
A convex polygon has 8 vertices. A diagonal is a segment joining two vertices that are not adjacent. How many diagonals does the polygon have? Enter your answer as a number.
Numeric entry — write the number.
Q13 · STA-CMB-025 · Select all that apply · 80s
A set S has exactly n distinct elements, where n >= 1. Which of the following are true for every such n? Select all that apply.
(A) The total number of subsets of S is 2^n
(B) C(n, 0) + C(n, 1) + ... + C(n, n) = 2^n
(C) The number of 2-element subsets of S is n(n+1)/2
(D) C(n, 1) = n
(E) C(n, n-1) = n
Level 3 · Hard — 12 questions · ~19 min
where 162+ is won or lost
Q14 · STA-CMB-039 · MCQ · 105s
A group of 10 people consists of 5 married couples. A committee of 4 is to be chosen; the members hold no distinct roles, so order does not matter, and no person may be chosen twice. In how many ways can the committee be chosen if no two people who are married to each other both serve on it?
(A) 70
(B) 80
(C) 130
(D) 210
(E) 1920
Q15 · STA-CMB-041 · MCQ · 95s
For a certain integer n >= 3, C(n, 3) = 4 x C(n, 2). What is the value of n?
(A) 6
(B) 12
(C) 13
(D) 14
(E) 16
Q16 · STA-CMB-045 · MCQ · 105s
A panel of 6 judges must be selected from 8 judges experienced in criminal law and 5 judges experienced in civil law. All 13 judges are distinct and no judge has both kinds of experience. The panel must include at least 4 criminal-law judges. Panel members hold no distinct roles, so order does not matter. How many different panels are possible?
(A) 700
(B) 980
(C) 1008
(D) 1716
(E) 2520
Q17 · STA-CMB-048 · MCQ · 105s
A school club has 15 students. Exactly 6 of them are in the band and exactly 5 are in the choir, and exactly 4 students are in both the band and the choir (those 4 are counted inside both the 6 and the 5). A group of 4 students is to be chosen; order does not matter and no student may be chosen twice. How many such groups contain at least 1 student who is in both the band and the choir?
(A) 330
(B) 1035
(C) 1295
(D) 1365
(E) 1456
Q18 · STA-CMB-050 · MCQ · 90s
A committee of 4 is to be chosen from a group of 9 distinct people. The members hold no distinct roles, so order does not matter. Two of the nine, Alice and Bob, refuse to serve together, so no committee may contain both of them. How many committees are possible?
(A) 70
(B) 84
(C) 105
(D) 119
(E) 126
Q19 · STA-CMB-046 · QC · 85s
A committee of 3 is to be chosen from 8 distinct people, one of whom is person A. The members hold no distinct roles, so order does not matter.
Column A: The number of such committees that include person A Column B: The number of such committees that exclude person A
Q20 · STA-CMB-051 · QC · 110s
A 5-card hand is dealt from a standard 52-card deck containing 13 hearts and 39 non-hearts. The order of the cards within a hand does not matter.
Column A: The number of hands containing exactly 2 hearts Column B: The number of hands containing exactly 3 hearts
Q21 · STA-CMB-044 · Numeric Entry · 85s
A group of 12 students contains 5 members of the debate club and 7 non-members. A team of 5 students is to be selected; the team members hold no distinct roles, so order does not matter, and no student may be selected twice. How many teams contain exactly 3 debate club members? Enter your answer as a number.
Numeric entry — write the number.
Q22 · STA-CMB-022 · Numeric Entry · 100s
A basket contains 6 red balls and 5 blue balls; all 11 balls are distinguishable. Four balls are selected at once, so order does not matter and no ball may be selected twice. In how many ways can the selection contain at least 1 red ball AND at least 1 blue ball? Enter your answer as a number.
Numeric entry — write the number.
Q23 · STA-CMB-036 · Numeric Entry · 90s
A set has 8 distinct elements. How many of its subsets contain at least 3 elements? Enter your answer as a number.
Numeric entry — write the number.
Q24 · STA-CMB-047 · Select all that apply · 85s
Which of the following pairs of values are equal? Select all that apply.
(A) C(15, 6) and C(15, 9)
(B) C(20, 8) and C(20, 12)
(C) C(10, 3) and C(10, 8)
(D) C(7, 2) and C(7, 5)
(E) C(n, 0) and C(n, n), for any positive integer n
Q25 · STA-CMB-061 · Select all that apply · 100s
In a league of n teams (n >= 2), every team plays every other team exactly once, and a game between two teams counts once regardless of which team is named first. Let G be the total number of games played. Which of the following must be true? Select all that apply.
(A) G = n(n-1)/2
(B) If n = 15, then G = 105
(C) If G = 45, then n = 10
(D) Doubling the number of teams doubles G
(E) G = P(n, 2)
Level 4 · Extreme — 5 questions · ~10 min
165+ — expect to need the insight, not the grind
Q26 · STA-CMB-054 · MCQ · 120s
A committee of 5 is to be chosen from a pool of 4 doctors, 3 lawyers and 2 engineers; all 9 people are distinct and each belongs to exactly one profession. The committee must contain at least 1 doctor and at least 1 lawyer. Committee members hold no distinct roles, so order does not matter. How many such committees are possible?
(A) 119
(B) 120
(C) 125
(D) 126
(E) 420
Q27 · STA-CMB-057 · MCQ · 115s
Ten distinct people are to be divided into one group of 4 and one group of 6. The two groups are distinguishable by their sizes, and within each group order does not matter. Three of the ten people are close friends who insist that all three of them end up in the same group. In how many ways can the division be made?
(A) 7
(B) 35
(C) 42
(D) 84
(E) 210
Q28 · STA-CMB-056 · QC · 125s
n is an integer greater than or equal to 3.
Column A: C(2n, n) Column B: 2 x C(2n-1, n-1)
Q29 · STA-CMB-055 · Numeric Entry · 110s
Six distinct people are to be divided into two groups of 3. The groups are unlabelled: the split {A, B, C} with {D, E, F} is the same division as {D, E, F} with {A, B, C}, and order within a group does not matter. In how many different ways can the division be made? Enter your answer as a number.
Numeric entry — write the number.
Q30 · STA-CMB-059 · Select all that apply · 120s
Let f(n) = C(2n, n) for positive integers n. Which of the following are true? Select all that apply.
(A) f(1) = 2
(B) f(2) = 6
(C) f(n+1) = 4 x f(n) for every n >= 1
(D) f(n) = 2 x C(2n-1, n) for every n >= 1
(E) f(3) = 15
Part B — Answer Key
| Q | ID | Level | Type | Answer |
|---|---|---|---|---|
| 1 | STA-CMB-006 |
easy | MCQ | B |
| 2 | STA-CMB-009 |
easy | MCQ | D |
| 3 | STA-CMB-011 |
easy | QC | B |
| 4 | STA-CMB-013 |
easy | Numeric Entry | 792 |
| 5 | STA-CMB-015 |
easy | Select all that apply | A, B, C, E |
| 6 | STA-CMB-016 |
medium | MCQ | C |
| 7 | STA-CMB-019 |
medium | MCQ | C |
| 8 | STA-CMB-029 |
medium | MCQ | C |
| 9 | STA-CMB-024 |
medium | QC | C |
| 10 | STA-CMB-034 |
medium | QC | C |
| 11 | STA-CMB-018 |
medium | Numeric Entry | 21 |
| 12 | STA-CMB-030 |
medium | Numeric Entry | 20 |
| 13 | STA-CMB-025 |
medium | Select all that apply | A, B, D, E |
| 14 | STA-CMB-039 |
hard | MCQ | B |
| 15 | STA-CMB-041 |
hard | MCQ | D |
| 16 | STA-CMB-045 |
hard | MCQ | C |
| 17 | STA-CMB-048 |
hard | MCQ | B |
| 18 | STA-CMB-050 |
hard | MCQ | C |
| 19 | STA-CMB-046 |
hard | QC | B |
| 20 | STA-CMB-051 |
hard | QC | A |
| 21 | STA-CMB-044 |
hard | Numeric Entry | 210 |
| 22 | STA-CMB-022 |
hard | Numeric Entry | 310 |
| 23 | STA-CMB-036 |
hard | Numeric Entry | 219 |
| 24 | STA-CMB-047 |
hard | Select all that apply | A, B, D, E |
| 25 | STA-CMB-061 |
hard | Select all that apply | A, B, C |
| 26 | STA-CMB-054 |
extreme hard | MCQ | A |
| 27 | STA-CMB-057 |
extreme hard | MCQ | C |
| 28 | STA-CMB-056 |
extreme hard | QC | C |
| 29 | STA-CMB-055 |
extreme hard | Numeric Entry | 10 |
| 30 | STA-CMB-059 |
extreme hard | Select all that apply | A, B, D |
Part C — Worked Solutions
Q1 · STA-CMB-006 — Answer: B
Step 1 — decide P or C. The two toppings sit on one pizza, so mushroom-then-olive is the same pizza as olive-then-mushroom. Order does not matter, so this is a combination. Step 2 — apply C(n,2) = n(n-1)/2. C(8,2) = (8 x 7)/2 = 56/2 = 28. Check by listing the pattern: topping 1 pairs with 7 others, topping 2 with 6 new ones, and so on: 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28. Correct. Answer: 28.
Trap. Answering 56, which is P(8,2) = 8 x 7 - the count of ORDERED pairs. That counts every pizza twice, once for each order of naming the toppings, so you must divide by 2! = 2. Choice E, 64 = 8^2, comes from allowing both order and repetition; choice C, 36 = C(9,2), is an off-by-one on n.
Q2 · STA-CMB-009 — Answer: D
Step 1 — every element is independently either in a subset or out of it, which is 2 choices per element. Step 2 — with 4 elements that gives 2 x 2 x 2 x 2 = 2^4 = 16 subsets. Step 3 — the same answer as a sum of combinations: C(4,0) + C(4,1) + C(4,2) + C(4,3) + C(4,4) = 1 + 4 + 6 + 4 + 1 = 16. Correct. Answer: 16.
Trap. Choice C, 15, comes from subtracting the empty set - but the question explicitly counts it. Choice E, 24 = 4!, comes from arranging the 4 elements instead of selecting subsets of them; a subset has no internal order at all.
Q3 · STA-CMB-011 — Answer: B
Column A - order does not matter, so this is a combination. C(5,3) = (5 x 4 x 3)/(3 x 2 x 1) = 60/6 = 10. Column B - the slots are labelled, so the same 3 books in different slots are different displays. This is a permutation. P(5,3) = 5 x 4 x 3 = 60. 60 > 10, so Column B is greater. The general link: P(n,r) = C(n,r) x r!, so Column B = Column A x 3! = 10 x 6 = 60. A permutation count is never smaller than the matching combination count.
Trap. Reading both columns as 'pick 3 from 5' and choosing C. The words that decide it are 'order does not matter' versus 'labelled slots'. Whenever the r selected items get distinct roles or positions, multiply the combination by r! to get the permutation.
Q4 · STA-CMB-013 — Answer: 792
Step 1 — order does not matter and no repetition is allowed, so this is C(12,5). Step 2 — write the formula and cancel before multiplying. C(12,5) = (12 x 11 x 10 x 9 x 8)/(5 x 4 x 3 x 2 x 1) Cancel: 12/(4 x 3) = 1, and 10/5 = 2. What is left: 11 x 2 x 9 x 8 / 2 = 11 x 9 x 8 = 792. Step 3 — sanity check with the symmetry rule: C(12,5) = C(12,7), and both must be whole numbers less than P(12,5) = 95040. 792 passes. Answer: 792.
Trap. Computing P(12,5) = 12 x 11 x 10 x 9 x 8 = 95040 and stopping there. That treats the five starters as an ordered list; since they hold identical roles you must divide by 5! = 120, and 95040/120 = 792.
Q5 · STA-CMB-015 — Answer: A, B, C, E
First compute the target: C(8,3) = (8 x 7 x 6)/(3 x 2 x 1) = 336/6 = 56. A - symmetry rule C(n,r) = C(n, n-r), and 8 - 3 = 5, so C(8,5) = C(8,3) = 56. TRUE. B - this is the definition n!/[r!(n-r)!] with n = 8, r = 3, since 8 - 3 = 5. It equals 56. TRUE. C - (8 x 7 x 6)/6 = 336/6 = 56, the cancelled form of the same formula (the 6 in the denominator is 3!). TRUE. D - 8 x 7 x 6 = 336 = P(8,3), the ordered count. It is 3! = 6 times too big. FALSE. E - Pascal's identity: C(7,2) + C(7,3) = 21 + 35 = 56. TRUE. Answer: A, B, C, E.
Trap. Selecting D because 8 x 7 x 6 is where every combination computation starts. That numerator is the permutation count; the division by 3! is what turns ordered selections into unordered ones. The other slip is rejecting A - students expect C(8,5) to be larger than C(8,3) because 5 > 3, but the symmetry rule makes them identical.
Q6 · STA-CMB-016 — Answer: C
Step 1 — the committee is built from two separate pools, so choose from each pool and MULTIPLY. Step 2 — men: C(6,2) = (6 x 5)/2 = 15. Step 3 — women: C(4,2) = (4 x 3)/2 = 6. Step 4 — total = 15 x 6 = 90. Check the size: 90 is well under C(10,4) = 210, the number of committees with no composition requirement. Correct. Answer: 90.
Trap. Choice A, 21, comes from ADDING the pools (15 + 6) instead of multiplying - but each of the 15 pairs of men can be paired with each of the 6 pairs of women, so the counts multiply. Choice B, 60, is C(6,2) x C(4,1) = 15 x 4, from choosing only one woman. Choice E, 210 = C(10,4), ignores the 2-and-2 requirement altogether, and choice D, 180, is P(6,2) x C(4,2) = 30 x 6, from ordering the two men.
Q7 · STA-CMB-019 — Answer: C
Method 1 - sum the cases by how many come from Section A. There are three cases: 2, 3 or 4 from A. Exactly 2 from A: C(5,2) x C(4,2) = 10 x 6 = 60. Exactly 3 from A: C(5,3) x C(4,1) = 10 x 4 = 40. Exactly 4 from A: C(5,4) x C(4,0) = 5 x 1 = 5. Total = 60 + 40 + 5 = 105. Method 2 - complement, which is faster here. Total selections = C(9,4) = 126. The forbidden cases are 0 or 1 from Section A: 0 from A: C(5,0) x C(4,4) = 1 x 1 = 1. 1 from A: C(5,1) x C(4,3) = 5 x 4 = 20. 126 - (1 + 20) = 126 - 21 = 105. Both methods agree. Answer: 105.
Trap. Choice B, 100, is 60 + 40 - the classic 'at least' error of summing the cases and forgetting the last one (all 4 from Section A). Choice E, 126, is C(9,4) with the constraint ignored, and choice D, 125, is 126 - 1, subtracting only the zero-from-A case while forgetting the 20 selections with exactly one.
Q8 · STA-CMB-029 — Answer: C
Step 1 — choose the finance committee from all 12: C(12,3) = (12 x 11 x 10)/6 = 220. Step 2 — those 3 are used up, so the social committee is chosen from the remaining 9: C(9,4) = (9 x 8 x 7 x 6)/24 = 3024/24 = 126. Step 3 — the two choices are made in sequence, so multiply: 220 x 126 = 27720. Check by reversing the order: C(12,4) x C(8,3) = 495 x 56 = 27720. Same answer, as it must be - the committees are distinguishable by name, so the order in which you fill them does not change the count. Answer: 27720.
Trap. Choice D, 108900 = C(12,3) x C(12,4), is the live error: choosing the second committee from all 12 again, which lets people sit on both committees when the stem forbids it. Choices A and B stop after one committee. Choice E, 3991680 = P(12,3) x P(9,4), treats each committee as an ordered list of office-holders.
Q9 · STA-CMB-024 — Answer: C
Column A: C(n,2) = n(n-1)/2. Column B: C(n-1,2) + (n-1) = (n-1)(n-2)/2 + (n-1). Factor out (n-1)/2 from Column B: (n-1)/2 x [(n-2) + 2] = (n-1)/2 x n = n(n-1)/2. That is exactly Column A, for every n > 2, so the two quantities are equal. The counting reason (Pascal's identity, C(n,r) = C(n-1,r) + C(n-1,r-1)): to pick 2 people from n, either both come from the first n-1 people, which is C(n-1,2) ways, or one of them is the nth person and the partner is any of the other n-1, which is n-1 ways. Check with n = 5: Column A = C(5,2) = 10; Column B = C(4,2) + 4 = 6 + 4 = 10. Equal. Test n = 8 too: 28 versus 21 + 7 = 28. Equal. Answer: C.
Trap. Choosing D because n is unknown. The identity holds for EVERY integer n > 2, so the variable never gets a chance to make the columns differ - plug in two different values of n and both give equality. The other error is expecting Column B to be smaller because it starts from C(n-1,2); the added (n-1) is exactly the shortfall.
Q10 · STA-CMB-034 — Answer: C
Column A: C(10,3) = (10 x 9 x 8)/(3 x 2 x 1) = 720/6 = 120. Column B: C(10,7) = C(10,3) = 120 by the symmetry rule. The reason is a one-to-one match: naming the 3 who serve automatically names the 7 who do not, and naming the 7 who do not automatically names the 3 who do. Every choice in Column A corresponds to exactly one choice in Column B, so the counts must be equal. That is the identity C(n,r) = C(n, n-r), with 3 + 7 = 10. Answer: C.
Trap. Assuming Column B is larger because 7 > 3, and choosing B. Choosing which people to leave out is the same act as choosing which people to take. When r is large, convert with the symmetry rule and compute the smaller one: C(10,7) should always be evaluated as C(10,3).
Q11 · STA-CMB-018 — Answer: 21
Step 1 — lock in the forced members. Priya and Raj take 2 of the 4 seats, and there is only 1 way to place them since order does not matter. Step 2 — that leaves 4 - 2 = 2 seats to fill from the 9 - 2 = 7 remaining candidates. Step 3 — C(7,2) = (7 x 6)/2 = 21. Check: total delegations with no restriction = C(9,4) = 126, and the fraction that contain both Priya and Raj should be small. 21/126 = 1/6, which is plausible. Correct. Answer: 21.
Trap. Computing C(9,2) = 36 by remembering to reduce the number of seats but forgetting to remove Priya and Raj from the candidate pool - they cannot be chosen again. The other error is C(7,4) = 35, which reduces the pool but not the number of open seats.
Q12 · STA-CMB-030 — Answer: 20
Step 1 — every segment between two vertices is an unordered pair of vertices, so the total number of segments is the handshake count C(8,2) = (8 x 7)/2 = 28. Step 2 — 8 of those segments are the sides of the polygon, and sides are not diagonals. Step 3 — diagonals = 28 - 8 = 20. General formula: n(n-3)/2 = 8 x 5/2 = 20. Correct. Answer: 20.
Trap. Answering 28, which counts every vertex pair and so includes the 8 sides. The second error is treating the segment from vertex 1 to vertex 5 as different from the segment from vertex 5 to vertex 1 - a segment is an unordered pair, so P(8,2) = 56 double counts everything.
Q13 · STA-CMB-025 — Answer: A, B, D, E
A - each element is independently in or out, giving 2^n subsets. TRUE. B - the left side counts subsets by size (size 0, size 1, ... , size n), and the right side counts them all at once, so the two must agree. Test n = 4: 1 + 4 + 6 + 4 + 1 = 16 = 2^4. TRUE. C - the number of 2-element subsets is C(n,2) = n(n-1)/2, not n(n+1)/2. Test n = 5: C(5,2) = 10, but 5 x 6/2 = 15. FALSE. D - there are exactly n ways to pick one element. TRUE. E - by the symmetry rule C(n, n-1) = C(n, 1) = n; picking n-1 elements is the same as picking the single element left out. TRUE. Answer: A, B, D, E.
Trap. C is the live error: the pairing formula is n(n-1)/2, and writing n(n+1)/2 (the sum-of-first-n-integers formula) gives 15 instead of 10 for a 5-element set. The two formulas look almost identical, so check them on a small case before trusting either.
Q14 · STA-CMB-039 — Answer: B
Direct method - build the committee couple by couple. Step 1 — the 4 members must come from 4 DIFFERENT couples, so first choose which 4 of the 5 couples are represented: C(5,4) = 5. Step 2 — from each chosen couple, take exactly one of the 2 spouses: 2 x 2 x 2 x 2 = 2^4 = 16. Step 3 — multiply: 5 x 16 = 80. Complement method as a check. Total committees = C(10,4) = 210. Committees containing at least one married pair: choose the pair, C(5,1) = 5, then 2 more from the remaining 8 people, C(8,2) = 28, giving 5 x 28 = 140 - but this counts each committee made of TWO complete couples twice, and there are C(5,2) = 10 of those. So committees with at least one pair = 140 - 10 = 130, and 210 - 130 = 80. Both methods give 80. Correct. Answer: 80.
Trap. Choice A, 70, is the complement done carelessly: 210 - 140, forgetting that the 140 double counts the 10 committees made of two whole couples. Choice C, 130, answers the opposite question - how many committees DO contain a married pair. Choice D, 210, is C(10,4) with the restriction dropped, and choice E, 1920 = 80 x 4!, orders the four members.
Q15 · STA-CMB-041 — Answer: D
Step 1 — write both sides in cancelled form. C(n,3) = n(n-1)(n-2)/6 and C(n,2) = n(n-1)/2. Step 2 — take the ratio rather than expanding. Divide C(n,3) by C(n,2): C(n,3)/C(n,2) = [n(n-1)(n-2)/6] x [2/(n(n-1))] = (n-2)/3. The n(n-1) cancels completely - this is the whole trick. Step 3 — the equation becomes (n-2)/3 = 4. Step 4 — n - 2 = 12, so n = 14. Check: C(14,3) = (14 x 13 x 12)/6 = 2184/6 = 364, and C(14,2) = (14 x 13)/2 = 91. Is 364 = 4 x 91? Yes, 4 x 91 = 364. Correct. Answer: 14.
Trap. Choice B, 12, is stopping at n - 2 = 12 and reporting that value as n. Choice A, 6, comes from solving n - 2 = 4, dropping the 3 in the denominator. Choice E, 16, comes from writing the ratio as (n-4)/3 by mishandling the (n-3)! term. The safeguard is the general fact C(n,r+1)/C(n,r) = (n-r)/(r+1): here (n-2)/3.
Q16 · STA-CMB-045 — Answer: C
'At least 4 criminal' allows 4, 5 or 6 criminal-law judges. There are only three cases, so summing is faster than the complement here. Case 1 - 4 criminal, 2 civil: C(8,4) x C(5,2) = 70 x 10 = 700. Case 2 - 5 criminal, 1 civil: C(8,5) x C(5,1) = 56 x 5 = 280. Case 3 - 6 criminal, 0 civil: C(8,6) x C(5,0) = 28 x 1 = 28. Total = 700 + 280 + 28 = 1008. Check: 1008 is below the unrestricted total C(13,6) = 1716, as it must be. Correct. Answer: 1008.
Trap. Choice E, 2520, is the seductive shortcut: pick 4 criminal judges first, C(8,4) = 70, then fill the last 2 seats from anyone left, C(9,2) = 36, giving 2520. That over-counts badly, because a panel with 5 criminal judges gets counted once for each choice of which 4 were the 'required' ones. Choice B, 980 = 700 + 280, drops the all-criminal case; choice A, 700, keeps only the first case; choice D, 1716 = C(13,6), ignores the requirement.
Q17 · STA-CMB-048 — Answer: B
Step 1 — identify the target set. 'In both' is exactly 4 students. The other 15 - 4 = 11 students are not in both. Step 2 — 'at least 1' is a signal to use the complement. Total groups of 4 from 15: C(15,4) = (15 x 14 x 13 x 12)/24 = 32760/24 = 1365. Step 3 — groups with NO 'both' student are chosen entirely from the other 11: C(11,4) = (11 x 10 x 9 x 8)/24 = 7920/24 = 330. Step 4 — at least 1 = 1365 - 330 = 1035. Check by summing cases: exactly 1: C(4,1)C(11,3) = 4 x 165 = 660; exactly 2: C(4,2)C(11,2) = 6 x 55 = 330; exactly 3: C(4,3)C(11,1) = 4 x 11 = 44; exactly 4: C(4,4) = 1. Sum = 660 + 330 + 44 + 1 = 1035. Correct. Answer: 1035.
Trap. Choice E, 1456 = C(4,1) x C(14,3) = 4 x 364, is the 'reserve one seat' over-count: a group containing 2 of the 4 both-students is counted twice, once for each of them in the reserved seat. Choice A, 330, is the complement itself - the answer to 'how many groups contain none'. Choice C, 1295 = 1365 - C(8,4), answers a different question, 'at least one student in band OR choir', since 15 - (6 + 5 - 4) = 8 students are in neither.
Q18 · STA-CMB-050 — Answer: C
Complement method. Step 1 — total committees with no restriction: C(9,4) = (9 x 8 x 7 x 6)/24 = 3024/24 = 126. Step 2 — committees that DO contain both Alice and Bob: lock them in, then fill the other 2 seats from the remaining 7 people: C(7,2) = 21. Step 3 — subtract: 126 - 21 = 105. Case check. Neither Alice nor Bob: C(7,4) = 35. Alice but not Bob: C(7,3) = 35. Bob but not Alice: C(7,3) = 35. Total 35 + 35 + 35 = 105. Correct. Answer: 105.
Trap. Choice A, 70, is the case sum with one case dropped - students count 'neither' and 'Alice only' but forget 'Bob only'. Choice B, 84, is 126 - 42, subtracting the 21 forbidden committees twice (once 'for Alice' and once 'for Bob'), but those are the same 21 committees. Choice D, 119 = 126 - 7, removes C(7,1) instead of C(7,2), forgetting that two seats remain after Alice and Bob are seated. Choice E, 126, ignores the restriction.
Q19 · STA-CMB-046 — Answer: B
Column A - person A takes one seat, so 2 seats remain and they are filled from the other 7 people: C(7,2) = (7 x 6)/2 = 21. Column B - person A is out, so all 3 seats are filled from the other 7: C(7,3) = (7 x 6 x 5)/6 = 210/6 = 35. 35 > 21, so Column B is greater. Check that the two columns partition everything: 21 + 35 = 56 = C(8,3), the total number of 3-person committees. Correct. The structural reason: including A costs a seat, leaving fewer ways to finish; excluding A costs a person but leaves all 3 seats free, and with 8 people and a committee this small the extra seat is worth more than the extra person.
Trap. Choosing C on the feeling that 'in' and 'out' are symmetric. They are not: the two counts are C(7,2) = 21 and C(7,3) = 35, and they are only equal when the numbers happen to make C(n-1, r-1) = C(n-1, r), which needs n = 2r. Here n = 8 and r = 3, so 2r = 6 is not 8 and the columns differ. Note the answer is not D either - both columns are fully determined numbers.
Q20 · STA-CMB-051 — Answer: A
Each hand splits into hearts and non-hearts, so each column is a product of two combinations. Column A = C(13,2) x C(39,3) = 78 x 9139. Column B = C(13,3) x C(39,2) = 286 x 741. Do not multiply these out - compare by ratio. Column A / Column B = [C(13,2)/C(13,3)] x [C(39,3)/C(39,2)] C(13,2)/C(13,3) = 78/286 = 3/11. C(39,3)/C(39,2) = [39 x 38 x 37/6] / [39 x 38/2] = 37/3. Ratio = (3/11) x (37/3) = 37/11, which is about 3.4 and clearly greater than 1. So Column A is greater. (The exact values are 712842 and 211926.) The intuition: only about a quarter of the deck is hearts, so hands lean toward having FEWER hearts; moving from 2 hearts to 3 hearts trades a plentiful non-heart for a scarce heart.
Trap. Choosing C because both columns 'use the same numbers' - C(13,2)C(39,3) and C(13,3)C(39,2) look like mirror images, but the two pools have very different sizes, so swapping which pool supplies the extra card changes the count by a factor of about 3.4. Choosing B by reasoning that 3 hearts is 'a stronger hand' confuses rarity with count: the rarer event has the SMALLER count.
Q21 · STA-CMB-044 — Answer: 210
Step 1 — 'exactly 3 debate members' fixes the split: 3 come from the 5 debate members and the other 5 - 3 = 2 come from the 7 non-members. Step 2 — debate members: C(5,3) = C(5,2) = (5 x 4)/2 = 10 (symmetry rule used to make r smaller). Step 3 — non-members: C(7,2) = (7 x 6)/2 = 21. Step 4 — the two pools are independent, so multiply: 10 x 21 = 210. Check the size: the unrestricted total is C(12,5) = 792, and 210 is a reasonable share of it. Correct. Answer: 210.
Trap. Choosing 3 debate members and then filling the last 2 seats from all 9 remaining students, C(5,3) x C(9,2) = 10 x 36 = 360. That allows a fourth debate member to slip into the team, which the word 'exactly' forbids; once the split is fixed at 3 and 2, the remaining seats must come only from the non-member pool.
Q22 · STA-CMB-022 — Answer: 310
Two conditions, so take the complement of each and subtract both. Step 1 — total selections of 4 from 11: C(11,4) = (11 x 10 x 9 x 8)/24 = 7920/24 = 330. Step 2 — selections with NO blue ball (all 4 red): C(6,4) = C(6,2) = 15. Step 3 — selections with NO red ball (all 4 blue): C(5,4) = C(5,1) = 5. Step 4 — those two bad families cannot overlap: a selection of 4 cannot be all-red and all-blue at once, so there is nothing to add back. 330 - 15 - 5 = 310. Check by summing the allowed splits (red, blue): (3,1) C(6,3)C(5,1) = 20 x 5 = 100; (2,2) C(6,2)C(5,2) = 15 x 10 = 150; (1,3) C(6,1)C(5,3) = 6 x 10 = 60. Total = 100 + 150 + 60 = 310. Correct. Answer: 310.
Trap. Subtracting only one complement and answering 315 (330 - 15) or 325 (330 - 5). Two separate 'at least one' conditions each rule out their own family of selections, and both must be removed. The other error is over-correcting by adding back an overlap of 'all red and all blue', which is empty here - inclusion-exclusion only adds a term back when the two bad families can actually share a member.
Q23 · STA-CMB-036 — Answer: 219
Step 1 — the total number of subsets of an 8-element set is 2^8 = 256. Step 2 — 'at least 3' has six cases (sizes 3 through 8), while its complement 'fewer than 3' has only three (sizes 0, 1, 2). Take the complement. C(8,0) = 1 (the empty set) C(8,1) = 8 C(8,2) = (8 x 7)/2 = 28 Complement total = 1 + 8 + 28 = 37. Step 3 — 256 - 37 = 219. Check by direct summation: C(8,3) + C(8,4) + C(8,5) + C(8,6) + C(8,7) + C(8,8) = 56 + 70 + 56 + 28 + 8 + 1 = 219. Correct. Answer: 219.
Trap. Forgetting the empty set and subtracting only 8 + 28 = 36, which gives 220. C(8,0) = 1: there is exactly one way to choose nothing, and it is a genuine subset with fewer than 3 elements. A second error is computing 2^8 as 8^2 = 64 or as 2 x 8 = 16 - the count of subsets is 2 raised to the number of elements.
Q24 · STA-CMB-047 — Answer: A, B, D, E
The test is the symmetry rule C(n,r) = C(n, n-r): two combinations with the same top number are equal exactly when their bottom numbers ADD UP to the top number. A - 6 + 9 = 15, which matches the top. Equal. TRUE. B - 8 + 12 = 20, which matches the top. Equal. TRUE. C - 3 + 8 = 11, but the top is 10, so the rule does not apply. Compute: C(10,3) = 120 and C(10,8) = C(10,2) = 45. Not equal. FALSE. D - 2 + 5 = 7, which matches the top. Equal (both are 21). TRUE. E - choosing none and choosing all are both single acts, so C(n,0) = C(n,n) = 1 for every positive integer n; also 0 + n = n. TRUE. Answer: A, B, D, E.
Trap. Selecting C because the numbers look like the others and 8 = 10 - 2 'feels' symmetric with 3. The partner of C(10,3) is C(10,7), not C(10,8). Always add the two lower numbers and check the sum against the upper number before declaring a pair equal.
Q25 · STA-CMB-061 — Answer: A, B, C
A game is an unordered pair of distinct teams, so G = C(n,2) = n(n-1)/2 - the handshake formula. A - that is the formula itself. TRUE. B - C(15,2) = (15 x 14)/2 = 210/2 = 105. TRUE. C - solve n(n-1)/2 = 45, so n(n-1) = 90. Since 10 x 9 = 90, n = 10. (The other root, n = -9, is rejected.) TRUE. D - test n = 5: G = C(5,2) = 10, and doubling to 10 teams gives C(10,2) = 45, not 20. In general C(2n,2) = n(2n-1), while 2 x C(n,2) = n(n-1), and these are equal only when 2n - 1 = n - 1, which is impossible for n >= 2. FALSE. E - P(n,2) = n(n-1) counts ORDERED pairs, which is exactly twice G. For n = 5 that is 20 rather than 10. FALSE. Answer: A, B, C.
Trap. E is the permutation-versus-combination trap in disguise: a game between Team X and Team Y is one game, not two, so the ordered count n(n-1) must be halved. D punishes linear thinking - the number of games grows roughly with the SQUARE of the number of teams, so doubling the league roughly quadruples the schedule (10 games becomes 45, not 20).
Q26 · STA-CMB-054 — Answer: A
Two 'at least one' conditions, so use inclusion-exclusion on the complement. Step 1 — total committees of 5 from 9: C(9,5) = C(9,4) = (9 x 8 x 7 x 6)/24 = 126. Step 2 — committees with NO doctor: choose all 5 from the 3 lawyers and 2 engineers, that is 5 people. C(5,5) = 1. Step 3 — committees with NO lawyer: choose all 5 from the 4 doctors and 2 engineers, that is 6 people. C(6,5) = 6. Step 4 — committees with neither a doctor nor a lawyer: all 5 would come from the 2 engineers, and C(2,5) = 0, so there is nothing to add back. Step 5 — 126 - 1 - 6 + 0 = 119. Check the structure: the only banned committees are the single all-lawyer-and-engineer committee and the six that use 4 doctors plus 1 engineer or 3 doctors plus 2 engineers - listing them gives C(4,3)C(2,2) = 4 and C(4,4)C(2,1) = 2, which is 6. Total banned = 7, and 126 - 7 = 119. Correct. Answer: 119.
Trap. Choice E, 420, is the reserve-a-seat over-count: pick 1 doctor (4 ways) and 1 lawyer (3 ways), then fill the remaining 3 seats from the other 7 people, C(7,3) = 35, giving 4 x 3 x 35 = 420. A committee with two doctors is then counted twice, once for each doctor in the 'reserved' seat. Choice B, 120, subtracts only the 6 no-lawyer committees; choice C, 125, subtracts only the 1 no-doctor committee; choice D, 126, is the unrestricted total.
Q27 · STA-CMB-057 — Answer: C
The whole division is determined once the group of 4 is named, since everyone else falls into the group of 6. So count the valid groups of 4. Case 1 - the 3 friends are all in the group of 4. Then 1 more seat is left and it is filled from the other 7 people: C(7,1) = 7. Case 2 - the 3 friends are all in the group of 6. Then the group of 4 contains none of them, so all 4 come from the other 7: C(7,4) = C(7,3) = 35. These cases cannot overlap - the friends are either all in the four or all in the six. Total = 7 + 35 = 42. Check: the unrestricted total is C(10,4) = 210, so 42 is exactly one fifth of all divisions. Correct. Answer: 42.
Trap. Choices A and B are the two cases taken alone - stopping after 'the friends are in the small group' gives 7, and stopping after 'the friends are in the big group' gives 35. Choice D, 84, doubles the answer by counting each division twice, as though naming the group of 6 were a separate outcome from naming the group of 4; it is not, because the two groups have different sizes and each choice of four fixes the six. Choice E, 210 = C(10,4), ignores the friends entirely.
Q28 · STA-CMB-056 — Answer: C
Step 1 — split Column A with Pascal's identity C(m, r) = C(m-1, r-1) + C(m-1, r), taking m = 2n and r = n: C(2n, n) = C(2n-1, n-1) + C(2n-1, n). Step 2 — apply the symmetry rule to the second term. In C(2n-1, n), the top is 2n-1 and (2n-1) - n = n-1, so C(2n-1, n) = C(2n-1, n-1). Step 3 — the two pieces are therefore identical, so C(2n, n) = C(2n-1, n-1) + C(2n-1, n-1) = 2 x C(2n-1, n-1), which is exactly Column B. This holds for every n, so the columns are equal. Numerical check, n = 3: Column A = C(6,3) = 20; Column B = 2 x C(5,2) = 2 x 10 = 20. And n = 4: C(8,4) = 70; 2 x C(7,3) = 2 x 35 = 70. Equal both times. Answer: C.
Trap. Choosing D because n is a variable, or choosing A on the feeling that C(2n, n) is 'much bigger' than a combination with a smaller top number - the factor of 2 in Column B closes the gap exactly. Testing one value of n and getting equality should push you to prove it rather than guess D; the identity is Pascal's rule plus symmetry, and it holds for every n.
Q29 · STA-CMB-055 — Answer: 10
Step 1 — first count as if the groups were labelled 'Group 1' and 'Group 2'. Choosing the 3 people for Group 1 fixes everything: C(6,3) = (6 x 5 x 4)/6 = 20. Step 2 — the groups are NOT labelled, so each division has been counted twice: once with a given trio as Group 1, and once with the same trio as Group 2. Step 3 — divide by 2: 20/2 = 10. Check by anchoring on one person. Fix person A; A's group is determined by which 2 of the other 5 people join A, which is C(5,2) = 10 - and this counts each division exactly once, because A lands in exactly one group. 10 confirms the answer. Answer: 10.
Trap. Answering 20, from C(6,3), which is the count when the two groups are distinguishable (say, 'the morning shift' and 'the evening shift'). The stem says the groups are unlabelled, so the division must be halved. The reverse error is over-dividing by 3! or by 2! twice - the only over-count here is the single swap of the two groups, so the divisor is 2.
Q30 · STA-CMB-059 — Answer: A, B, D
Compute the first few values: f(1) = C(2,1) = 2, f(2) = C(4,2) = 6, f(3) = C(6,3) = 20, f(4) = C(8,4) = 70. A - f(1) = C(2,1) = 2. TRUE. B - f(2) = C(4,2) = (4 x 3)/2 = 6. TRUE. C - test at n = 1: f(2) = 6, but 4 x f(1) = 4 x 2 = 8. The claim fails immediately. FALSE. (The true ratio is f(n+1)/f(n) = 2(2n+1)/(n+1), which is 3 at n = 1, then 10/3, then 3.5 - it only approaches 4 as n grows.) D - by Pascal's identity C(2n, n) = C(2n-1, n-1) + C(2n-1, n), and by symmetry C(2n-1, n-1) = C(2n-1, n) because (n-1) + n = 2n-1. So C(2n, n) = 2 x C(2n-1, n). Check n = 3: 2 x C(5,3) = 2 x 10 = 20 = f(3). TRUE. E - f(3) = C(6,3) = (6 x 5 x 4)/(3 x 2 x 1) = 120/6 = 20, not 15. FALSE. Answer: A, B, D.
Trap. E is the arithmetic bait: computing C(6,3) with the pairing formula n(n-1)/2 = (6 x 5)/2 = 15 instead of the three-item formula, which needs the extra factor 4/3. C is the pattern bait: the ratios 3, 3.33, 3.5, ... creep toward 4 without ever reaching it, so a student who checks only large n or trusts the 'doubling the top doubles the exponent' feeling will accept it.