GRE Permutations and Combinations Tips for Choosing Formulas and Avoiding Mistakes

If permutations and combinations feel like a tangle of formulas and exceptions, this article gives clear, high-impact tips you can use right away on the GRE. You will learn how to decide whether to count ordered arrangements or unordered selections, pick the correct formula, use shortcuts that save time, and avoid the common mistakes that cost points under test conditions.

Quick answer

Permutations count ordered arrangements; combinations count unordered selections. On the GRE, decide whether order matters, check whether repetition is allowed, then apply the correct formula: nPr = n! / (n – r)! for permutations, nCr = n! / (r! (n – r)!) for combinations. Use complementary counting, casework, symmetry, and cancellation to simplify factorials, and practice translating word problems into one counting question before computing.

Why this matters for GRE performance

Permutations and combinations problems appear frequently in the Quant section and in probability questions. A single misinterpretation of “order matters” versus “order does not matter” will turn a correct logical approach into a wrong numeric answer. Competent, fast handling of counting problems improves accuracy and saves time for harder questions.

Common challenges students face

• Confusing order and selection. Students often answer as if order matters when the test means only the set of chosen items matters, and vice versa.
• Forgetting repetition rules. Repetition allowed versus repetition not allowed changes the counting method entirely.
• Overusing factorials. Writing large factorials without cancelling leads to heavy calculation and arithmetic errors.
• Not using complementary counting. Some problems are easier when you count the complement and subtract from the total.
• Poor setup of multi-case problems. Mixed conditions often require clean casework; skipping cases causes undercounting or double counting.

Core formulas and a quick reference

Keep these formulas and notes memorized, but focus on meaning rather than rote memorization.

Step-by-step strategy to solve GRE counting problems

1. Read carefully and restate the question. Convert the sentence into a counting question. Ask yourself: “Am I counting ordered arrangements or unordered selections?” Write that down in one clear sentence.
2. Identify the universe and constraints. Determine n (total distinct options), r (how many chosen or placed), whether repetition is allowed, and any special restrictions (adjacency banned, identical items, circular seating, symmetry).
3. Choose the simplest model. Decide which formula fits directly. If none fits exactly, consider splitting into disjoint cases or using complementary counting.
4. Simplify before computing. Cancel factorials symbolically whenever possible. For example, compute nPr by writing n × (n – 1) × … × (n – r + 1) instead of evaluating factorials.
5. Use symmetry and complementary counting. If half of the outcomes are symmetric, use multiplication or division accordingly. If counting directly is messy, count the complement and subtract from the total.
6. Do clean casework when needed. Break the problem into mutually exclusive cases, count each, then add. Use labels so you do not double count overlapping cases.
7. Estimate to check plausibility. After an exact calculation, do a quick size check. If an answer is extremely large or small relative to n and r, re-check the model.
8. Use answer choices to reverse-check. On the GRE, plug choices back into a simplified reasoning test if you are unsure which counting model applies.

How to implement cancellation and reduce arithmetic

• Write permutations as products, for example 10P4 = 10 × 9 × 8 × 7. Cancel numerators and denominators before multiplying.
• When using combinations, expand r! in the denominator and cancel with consecutive terms in the numerator. For 12C5, write 12 × 11 × 10 × 9 × 8 / 5! and cancel factors with 5! = 120.
• Factor numbers to cancel common prime factors rather than multiply large numbers directly.

Common mistakes and how to avoid them

• Mistake: Treating unordered selections as ordered. Fix: Ask “Does ABC differ from CBA?” If no, use combinations.
• Mistake: Ignoring identical items. Fix: If items are identical, divide by factorials of identical counts.
• Mistake: Overcounting by overlapping cases. Fix: Ensure cases are mutually exclusive and exhaustive, or use inclusion-exclusion.
• Mistake: Using permutations when repetition is allowed but positions are treated as distinct. Fix: Use n^r for ordered repetition problems.
• Mistake: Skipping complementary counting because direct counting seems more “natural.” Fix: Always test the complement if direct setup is long or messy.

Practice and implementation: worked examples

Example 1. Simple selection versus arrangement

Question: From 8 students, how many ways can a committee of 3 be chosen?

Analysis: Order does not matter. Use combinations: 8C3 = 8 × 7 × 6 / 3 × 2 × 1 = 56.

Example 2. Ordered arrangements

Question: From the same 8 students, how many ways can a president, vice president, and secretary be chosen?

Analysis: Order matters because positions are distinct. Use permutations: 8P3 = 8 × 7 × 6 = 336.

Example 3. Repetition allowed, ordered

Question: How many 4-letter passwords can be made from 26 letters when letters may repeat?

Analysis: For each of 4 positions choose any of 26 letters. Use n^r: 26^4 = 456,976. This is faster than trying to build factorial expressions.

Example 4. Combinations with repetition

Question: How many ways can you choose 5 candies from 3 types if you can take any number of each type?

Analysis: Use stars and bars: (n + r – 1)Cr = (3 + 5 – 1)C5 = 7C5 = 21. Alternatively compute 7C2 = 21, since 7C5 = 7C2.

Example 5. Arrangements with identical items

Question: How many distinct letter arrangements are there for the word STATISTICS?

Analysis: The letters are S S S T T T A I I C. Total letters n = 10. Counts: S=3, T=3, A=1, I=2, C=1. Use multinomial formula: 10! / (3! 3! 2! 1! 1!) = 10! / (3! 3! 2!) = 3,628,800 / (6 × 6 × 2) = 3,628,800 / 72 = 50,400.

Example 6. Circular permutations

Question: In how many different ways can 6 friends sit around a round table?

Analysis: Rotations are identical. For distinct people around a circle, use (n – 1)!. So 5! = 120.

Example 7. Complementary counting to avoid messy casework

Question: How many 5-letter strings from letters A, B, C, D, E have at least one A, repetition allowed?

Analysis: Total strings = 5^5 = 3125. Strings with no A = 4^5 = 1024. So answer = 3125 – 1024 = 2101. Complementary counting is much faster than counting strings with at least one A directly.

Example 8. GRE-style combined conditions (worked with cases)

Question: From digits 1 through 6, how many 3-digit numbers can be formed with distinct digits and digits in increasing order?

Analysis: Distinct digits and increasing order means each chosen set of 3 digits corresponds to exactly one increasing-ordered number. So count unordered 3-element selections from 6, which is 6C3 = 20. This is a classic trick: when a set of chosen digits must be in increasing order, use combinations.

Timed-practice set

Solve each problem under 2 minutes, then check the answers provided below.

1. How many 4-letter sequences can be formed from the letters A, B, C, D if repetition is allowed?
2. How many ways to choose 3 books from 10 different books?
3. How many 5-digit numbers with distinct digits can be formed from digits 0-9 if the number cannot start with 0?
4. How many different arrangements of the letters in the word MISSISSIPPI exist?
5. From 7 people, how many ways can you select a president and a treasurer if the same person cannot hold both offices?

Answers and brief explanations

1. 26^4 = 4^4 = 256, wait correct base is 4 letters so 4^4 = 256, because repetition allowed and 4 choices per slot.
2. 10C3 = 120.
3. First choose the leading digit from 1-9 excluding 0, so choose from 1-9 but we only have 0-9 and need distinct digits, correct approach: choose leading digit from 1-9 giving 9 choices, then choose 4 distinct digits from the remaining 9 digits and arrange them in the remaining 4 positions without repetition. Better approach: Count permutations where first digit is nonzero: choose the first digit in 9 ways (1-9), then choose and arrange the remaining 4 digits from the remaining 9 digits in P(9,4) ways. So total = 9 × 9P4 = 9 × (9 × 8 × 7 × 6) = 9 × 3024 = 27,216.
4. MISSISSIPPI letters count: M=1, I=4, S=4, P=2; total letters n=11. Number = 11! / (1! 4! 4! 2!) = 34,650.
5. Ordered distinct offices: 7 choices for president times 6 remaining choices for treasurer = 42.

High-impact tips for test day

• Underline or rewrite the key phrase that tells you whether order matters. Typical trigger words for order matter: “arrange”, “order”, “permutation”, “line up”, “ordered”. Trigger words for no order: “choose”, “committee”, “select”, “subset”.
• When numbers are large, simplify symbolically. For example, reduce nCr by cancelling r! with part of n × (n – 1) × … rather than computing full factorials.
• Look for one-to-one correspondences. If each selection corresponds to a unique ordered object, you may use either permutations or combinations suitably, but pick the simpler.
• Prefer complement or case splitting when direct counting requires many overlapping conditions.
• Estimate order of magnitude to eliminate wildly wrong answer choices quickly on multiple choice questions.
• Memorize a few patterns: n^r for ordered repetition, nPr for ordered no repetition, nCr for unordered no repetition, stars and bars for unordered repetition.

Frequently asked questions

Q: How do I know when to use complementary counting?

A: Use the complement when the condition you want is “at least one” or when avoiding a restriction is simpler than counting allowed cases. For example “at least one fixed element” is usually complement-friendly. If the prohibited sets are few and simple, count the complement.

Q: When should I do casework versus inclusion-exclusion?

A: Do casework when distinct cases are easy to describe and small in number. Use inclusion-exclusion when cases overlap and the overlaps are manageable to compute. On GRE problems, simple inclusion-exclusion with two or three overlapping sets is common and efficient.

Q: Are there tricks for arrangements with identical items?

A: Yes. Start with n! then divide by factorials of identical counts. If the arrangement has additional structure, like fixed positions for some items, adjust the numerator and denominator accordingly. Also consider grouping identical items first if that simplifies counting.

Q: How to deal with problems that mix probability and counting?

A: Translate the probability into a ratio of counts: favorable outcomes over total possible outcomes. Carefully ensure numerator and denominator are modeled the same way, with the same assumptions about order and repetition.

Q: Should I memorize factorials beyond 10!?

A: No. Memorize small factorials like 3!, 4!, 5!, and be comfortable writing permutations as products to cancel. The GRE rarely requires computing very large factorials directly. Use cancellation and simplification instead.

Final thoughts

Improving on GRE permutations and combinations depends less on memorizing many formulas and more on developing a clear translation skill. Always restate the counting task, identify whether order and repetition matter, choose the simplest model, and simplify before you calculate. Practice translating word problems into one-line counting questions, and you will reduce errors and save time on test day.