Permutation & Combination
Permutation and Combination is the gateway to all of counting — and to Probability, which sits right next to it on every CAT and XAT paper. The entire chapter rests on one decision: does order matter? If rearranging the chosen items creates a genuinely new outcome (seats in a row, a 3-digit code, a podium of gold-silver-bronze), you are counting arrangements — permutations. If only the group matters and shuffling it changes nothing (a committee, a hand of cards, a salad of fruits), you are counting selections — combinations. Once that question is settled, two formulas do almost everything: nPr = n!/(n−r)! for arranging r of n, and nCr = n!/(r!(n−r)!) for choosing r of n. CAT rarely asks a clean "how many ways"; it buries the count inside restrictions — items that must sit together, items that must stay apart, repeated letters, people around a round table. This chapter builds the full toolkit: the fundamental counting principle, arrangements with repetition, circular and necklace arrangements, and the gap and grouping methods that crack restricted problems fast.
Topics
⚡ CAT shortcuts & speed methods
The fastest ways to crack this chapter under time pressure — the techniques that separate a 95+ percentiler from the rest.
- First decide: does order matter? Order ⇒ permutation (nPr); group only ⇒ combination (nCr). This one question solves half the chapter.
- Use nCr = nC(n−r) to compute the easier side: 10C8 is just 10C2 = 45.
- Draw blanks for digit/code problems and write valid choices per slot, then multiply — watch the leading-zero and repetition rules.
- Round table = (n−1)!; necklace/garland (can flip) = (n−1)!/2; numbered seats or a fixed head = n! (no free rotation).
- Together ⇒ glue into a block: (units)! × (internal)!. Apart ⇒ gap method: arrange the rest, then drop items into the gaps.
- Repeated letters ⇒ divide n! by the factorial of each repeat count (BANANA = 6!/(3!2!) = 60).
⚠️ Common mistakes & traps
CAT is designed so that careless errors here cost you marks. Internalise each trap before the exam.
- Using a permutation when order does not matter (counting a committee as if seats were ordered) — inflates the answer by r!.
- Forgetting the leading digit cannot be 0 in number-formation problems.
- Using n! for a round table instead of (n−1)!, or forgetting the extra ÷2 for a necklace/garland.
- Not dividing by the factorials of repeated letters (treating BANANA’s three A’s as distinct).
- Adding cases joined by "and" or multiplying cases joined by "or" — it is the reverse: "and" multiplies, "or" adds.
📈 CAT exam insight & PYQ analysis
🎴 Flashcards — instant recall
Tap a card to reveal the answer. Drill these until they are automatic.
📌 Quick revision
Chapter test
🏆 Vidaara CAT success checklist
You have truly mastered Permutation & Combination when you can tick every box below.
- Recall every formula in this chapter without looking them up
- Solve each topic’s practice set with at least 80% accuracy
- Use the chapter shortcuts to cut your solving time in half
- Spot and avoid every common trap listed above
- Score 80%+ on the timed chapter test
📋 Chapter mastery scorecard
Track where you stand. Aim for the target before moving to the next chapter.
| Skill checkpoint | Target |
|---|---|
| Concept theory & formulas understood | 100% |
| Topic practice sets attempted (4 topics) | 4/4 |
| Best topic-test score | — → 80%+ |
| Chapter test score | — → 80%+ |
| Flashcards drilled to instant recall | 12 cards |
Formula Reference Sheet
Core counting
| Permutations (order matters) | nPr = n! / (n − r)! |
|---|---|
| Combinations (order ignored) | nCr = n! / (r!(n − r)!) |
| Arrangements with repetition | n! / (p! q! r! …) for repeated items |
| Fundamental counting principle | total = (ways for step 1) × (ways for step 2) × … |
| Symmetry of combinations | nCr = nC(n − r) |
Circular & special
| Circular arrangement (n distinct) | (n − 1)! |
|---|---|
| Necklace / garland (reflections same) | (n − 1)! / 2 |
| Sum of all combinations | nC0 + nC1 + … + nCn = 2ⁿ |
| At least one selection (from n) | 2ⁿ − 1 |
| Relation P and C | nPr = nCr × r! |