GMAT Word Problems: Probability, Combinatorics
Last updated: May 2, 2026
Word Problems: Probability, Combinatorics questions are one of the highest-leverage areas to study for the GMAT. This guide breaks down the rule, the elements you need to recognize, the named traps that catch most students, and a memory aid that scales to test day. Read it once, then practice the same sub-topic adaptively in the app.
The rule
Probability is the count of outcomes you want divided by the count of total equally likely outcomes, so most probability problems collapse into two counting problems. Use permutations when the order of the chosen items matters and combinations when it does not, and use the multiplication principle to chain independent stages of a selection. When direct counting is messy, switch to the complement: $P(A) = 1 - P(\text{not } A)$.
Elements breakdown
Frame the Sample Space
Decide what counts as one outcome and whether outcomes are equally likely.
- Identify the single random act
- Decide if order distinguishes outcomes
- Confirm replacement or no replacement
- Confirm distinct or identical objects
Choose Permutation or Combination
Pick the counting tool that matches whether order matters.
- Order matters then use $P(n,k) = \frac{n!}{(n-k)!}$
- Order does not matter then use $C(n,k) = \frac{n!}{k!(n-k)!}$
- Arrangements of all $n$ distinct items use $n!$
- Identical items divide by repeats $\frac{n!}{a!b!\dots}$
Apply the Multiplication Principle
Chain independent stages by multiplying the count at each stage.
- Break the task into ordered stages
- Count options at each stage independently
- Multiply stage counts for total arrangements
- Adjust for restrictions before multiplying
Use Complement and Addition Rules
Combine event probabilities using union, intersection, and complement.
- At least one means $1 - P(\text{none})$
- Mutually exclusive events add: $P(A) + P(B)$
- Independent events multiply: $P(A) \cdot P(B)$
- Inclusion-exclusion: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Handle Restrictions and Groupings
Adjust counts when items must be together, apart, or in fixed positions.
- Treat must-be-together items as one block
- For not-adjacent: place others first then insert
- Fix forbidden positions then count remainder
- Subtract bad arrangements from unrestricted total
Common patterns and traps
The At-Least-One Complement
When a question asks for the probability of at least one success across multiple trials, computing it directly forces you to add several disjoint cases (exactly one, exactly two, and so on). The fast move is the complement: subtract the probability of zero successes from $1$. This pattern shows up whenever you see the phrases 'at least one,' 'at least once,' or 'none of them.'
An answer choice often hands you a clean fraction like $\frac{37}{40}$ or $1 - \left(\frac{2}{5}\right)^3$, while wrong choices give the raw 'zero successes' probability or the probability of exactly one success.
Order-Matters Mismatch
GMAT problems sometimes describe selections that sound like simple groupings but actually distinguish positions, such as president and treasurer versus a two-person committee. Solving with the wrong tool inflates or deflates the count by exactly $k!$. Always test whether swapping two of your picks would change the outcome.
The trap answer is the correct numerator multiplied or divided by $k!$ — for example, $20$ instead of $120$, or $720$ instead of $120$, depending on which direction you slipped.
The Identical-Items Overcount
When you arrange items where some are indistinguishable — like the letters of a word with repeats or scoops of the same flavor — using plain $n!$ overcounts. The fix is to divide by the factorials of each repeated group. Forgetting this is a top-three combinatorics error.
The wrong answer is the unadjusted $n!$ figure (for example $5040$ for arranging $7$ letters with repeats), while the correct answer is that figure divided by something like $2! \cdot 3!$.
Together-or-Apart Restriction
When the problem requires certain items to stand together, glue them into a single block, count arrangements of the resulting items, then multiply by the internal arrangements of the block. When the problem requires items to stand apart, place the unrestricted items first and slot the restricted items into the gaps. Mixing up these two procedures produces predictable trap answers.
Trap choices substitute the 'together' answer when the question said 'not together,' or vice versa, often differing by exactly the unrestricted total minus the together count.
Replacement Slip
Probability problems that draw items one at a time can be with or without replacement. Without replacement, the denominator shrinks each draw; with replacement, it stays constant. Students often default to one model when the problem specified the other.
Wrong answers commonly look like $\left(\frac{4}{10}\right)^2 = \frac{16}{100}$ when the right computation was $\frac{4}{10} \cdot \frac{3}{9} = \frac{12}{90}$, or the reverse.
How it works
Almost every GMAT probability or combinatorics problem reduces to a simple two-step move: figure out the size of the sample space, then figure out the size of the favorable subset. For instance, suppose a small consulting firm has $4$ analysts and $3$ associates and wants to pick a $3$-person team. The total number of teams is $C(7,3) = 35$. The number of teams with exactly one associate is $C(3,1) \cdot C(4,2) = 3 \cdot 6 = 18$, so the probability is $\frac{18}{35}$. Notice how order never mattered, so we used combinations, and how we split the favorable count into two independent picks and multiplied. When the question asks for at least one of something, your reflex should be the complement: counting everything that has zero of the thing is almost always faster than summing one, two, three, and so on.
Worked examples
A regional sales office has $5$ junior reps and $4$ senior reps. The office manager will randomly select a $3$-person task force from the $9$ reps. What is the probability that the task force contains at least $1$ senior rep?
What is the probability that the task force contains at least $1$ senior rep?
- A $\frac{5}{42}$
- B $\frac{10}{21}$
- C $\frac{1}{2}$
- D $\frac{31}{42}$
- E $\frac{37}{42}$ ✓ Correct
Why E is correct: Use the complement. The total number of $3$-person task forces is $C(9,3) = 84$. The number with zero senior reps (all juniors) is $C(5,3) = 10$, so $P(\text{no seniors}) = \frac{10}{84} = \frac{5}{42}$. Therefore $P(\text{at least one senior}) = 1 - \frac{5}{42} = \frac{37}{42}$.
Why each wrong choice fails:
- A: This is $P(\text{no senior reps})$, the probability you should subtract from $1$, not the final answer. The student stopped one step short. (The At-Least-One Complement)
- B: This equals $\frac{C(4,1) \cdot C(5,2)}{84} = \frac{40}{84}$, the probability of exactly $1$ senior. It ignores task forces with $2$ or $3$ seniors. (The At-Least-One Complement)
- C: This is the rough proportion of seniors in the office ($\frac{4}{9} \approx \frac{1}{2}$), not a probability built from counting favorable task forces.
- D: This is $\frac{62}{84}$, which arises from double counting: adding $P(\text{exactly 1}) + P(\text{exactly 2})$ but mishandling the exactly-$2$ count and missing the exactly-$3$ case. (The At-Least-One Complement)
A book club must arrange $7$ books on a shelf: $3$ identical copies of a marketing textbook, $2$ identical copies of a finance textbook, and $2$ different novels. In how many distinguishable ways can the $7$ books be arranged in a row?
In how many distinguishable ways can the $7$ books be arranged in a row?
- A $210$
- B $420$ ✓ Correct
- C $840$
- D $2520$
- E $5040$
Why B is correct: If all $7$ items were distinct, there would be $7! = 5040$ arrangements. Because $3$ of the marketing books are identical and $2$ of the finance books are identical, divide by $3!$ for the marketing repeats and by $2!$ for the finance repeats: $\frac{5040}{3! \cdot 2!} = \frac{5040}{12} = 420$. The two novels are distinct, so they introduce no further division.
Why each wrong choice fails:
- A: This divides by an extra $2!$, treating the two distinct novels as if they were identical. Distinct items should never be divided out. (The Identical-Items Overcount)
- C: This divides only by $3!$ for the marketing copies and forgets to divide by $2!$ for the identical finance copies. (The Identical-Items Overcount)
- D: This divides only by $2!$, treating only the finance pair as identical and ignoring the three identical marketing copies. (The Identical-Items Overcount)
- E: This is $7!$ unadjusted, which counts every identical book as if it were unique. It is the classic 'forgot to divide' trap. (The Identical-Items Overcount)
A jar contains $6$ red marbles and $4$ blue marbles. Two marbles are drawn at random, one after the other, without replacement. What is the probability that the two marbles drawn are different colors?
What is the probability that the two marbles drawn are different colors?
- A $\frac{6}{25}$
- B $\frac{12}{25}$
- C $\frac{4}{15}$
- D $\frac{8}{15}$ ✓ Correct
- E $\frac{1}{2}$
Why D is correct: Compute $P(\text{red then blue}) + P(\text{blue then red})$ without replacement: $\frac{6}{10} \cdot \frac{4}{9} + \frac{4}{10} \cdot \frac{6}{9} = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}$. Equivalently, $\frac{C(6,1) \cdot C(4,1)}{C(10,2)} = \frac{24}{45} = \frac{8}{15}$.
Why each wrong choice fails:
- A: This is $\frac{6}{10} \cdot \frac{4}{10} = \frac{24}{100}$, which uses replacement and only counts the red-then-blue order. It makes both errors at once. (Replacement Slip)
- B: This doubles the with-replacement single-order calculation $2 \cdot \frac{6}{10} \cdot \frac{4}{10} = \frac{48}{100}$. The doubling is right but the replacement assumption is wrong. (Replacement Slip)
- C: This is $\frac{6}{10} \cdot \frac{4}{9} = \frac{24}{90} = \frac{4}{15}$, the probability of red first then blue only. It forgets that blue first then red also gives different colors. (Order-Matters Mismatch)
- E: This is the rough idea that 'about half the marbles are each color, so different colors should be near $\frac{1}{2}$.' It is intuition, not a calculation.
Memory aid
OSCAR: Order? Sample space? Complement? Adjust for restrictions? Reduce the fraction.
Key distinction
Permutations count arrangements where position matters ($n!/(n-k)!$); combinations count groupings where position is irrelevant ($n!/(k!(n-k)!)$). If swapping two chosen items would create a different outcome, it is a permutation; otherwise it is a combination.
Summary
Reduce every probability question to two counts — favorable over total — and pick permutations, combinations, or the complement based on what the wording actually requires.
Practice word problems: probability, combinatorics adaptively
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Start your free 7-day trialFrequently asked questions
What is word problems: probability, combinatorics on the GMAT?
Probability is the count of outcomes you want divided by the count of total equally likely outcomes, so most probability problems collapse into two counting problems. Use permutations when the order of the chosen items matters and combinations when it does not, and use the multiplication principle to chain independent stages of a selection. When direct counting is messy, switch to the complement: $P(A) = 1 - P(\text{not } A)$.
How do I practice word problems: probability, combinatorics questions?
The fastest way to improve on word problems: probability, combinatorics is targeted, adaptive practice — working questions that focus on your specific weak spots within this sub-topic, getting immediate feedback, and revisiting items you missed on a spaced-repetition schedule. Neureto's adaptive engine does this automatically across the GMAT; start a free 7-day trial to see your sub-topic mastery climb in real time.
What's the most important distinction to remember for word problems: probability, combinatorics?
Permutations count arrangements where position matters ($n!/(n-k)!$); combinations count groupings where position is irrelevant ($n!/(k!(n-k)!)$). If swapping two chosen items would create a different outcome, it is a permutation; otherwise it is a combination.
Is there a memory aid for word problems: probability, combinatorics questions?
OSCAR: Order? Sample space? Complement? Adjust for restrictions? Reduce the fraction.
What's a common trap on word problems: probability, combinatorics questions?
Confusing order-matters with order-doesn't-matter
What's a common trap on word problems: probability, combinatorics questions?
Counting with replacement when there is none
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