Why the modelling is the difficulty
Consider two questions. How many ways can 3 people be chosen from 7 to form a committee? How many ways can 3 people be chosen from 7 to be president, secretary and treasurer? The numbers involved are identical and the answers differ by a factor of six, because the second distinguishes who holds which post and the first does not.
Every hard question in this chapter is a version of that distinction dressed differently - identical balls versus distinct balls, seats in a row versus seats around a table, letters that repeat versus letters that do not. Once the situation is modelled correctly the arithmetic is a single line, which is why practice should be about varied situations rather than harder computations.
How JEE actually asks Permutations and Combinations
NTA does not publish chapter-wise weightage, so figures circulating online are coaching estimates from past papers; check the official NTA information bulletin for your session's syllabus and pattern.
The recurring types are word-arrangement problems with repeated letters, selection problems with constraints such as at least one of a category, circular arrangements, and distribution of identical objects into distinct groups. The chapter also underpins Probability, where most questions are counting problems with a division at the end.
Key concepts, compressed
- ✓The fundamental counting principle: if one stage can happen in m ways and a second in n ways, the two together happen in mn ways.
- ✓A permutation counts arrangements, where order matters. A combination counts selections, where it does not.
- ✓ⁿPr = n!/(n − r)! and ⁿCr = n!/(r!(n − r)!), so ⁿPr = ⁿCr × r! - the extra r! is the number of orderings of each selection.
- ✓Circular arrangements of n distinct objects number (n − 1)!, since rotations of the same arrangement are not distinct.
- ✓Arrangements of objects with repeats divide by the factorial of each repeat count.
- ✓Identical objects into distinct boxes is a different problem entirely, solved by the stars-and-bars formula.
Formulas you need before attempting the questions
| Situation | Formula | Note |
|---|---|---|
| Arrangements of n distinct | n! | |
| r from n, order matters | ⁿPr = n!/(n − r)! | |
| r from n, order does not | ⁿCr = n!/(r!(n − r)!) | |
| Relation | ⁿPr = ⁿCr × r! | |
| Symmetry | ⁿCr = ⁿC(n − r) | |
| Circular arrangements | (n − 1)! | |
| Circular, reflections identical | (n − 1)!/2 | necklaces, garlands |
| With repeated objects | n!/(p! q! r! …) | |
| Identical into distinct boxes | ⁿ⁺ʳ⁻¹C(r−1) | boxes may be empty |
| Identical, each box non-empty | ⁿ⁻¹C(r−1) | |
| Total subsets of n elements | 2ⁿ | |
| Diagonals of an n-gon | ⁿC₂ − n | |
| 0! | 1 | by definition |
The five mistakes that cost the most marks
- ✓Using a permutation where order does not matter. Ask whether swapping two chosen items produces a different outcome before choosing the formula.
- ✓Forgetting to divide by the factorials of repeated letters. Arranging the letters of a word with repeats and using n! inflates the answer.
- ✓Treating a circular arrangement as n!. Fixing one position gives (n − 1)!; using n! counts each arrangement n times.
- ✓Adding where you should multiply. Multiply for stages that happen together, add for mutually exclusive cases.
- ✓Attacking 'at least one' directly. Counting the complement and subtracting from the total is faster and avoids missing a case.
Practice set 1: the counting principle
1. A person has 3 shirts and 4 pairs of trousers. How many outfits are possible?
12. Choosing a shirt and choosing trousers are two stages that happen together, so multiply: 3 × 4 = 12. This is the fundamental counting principle, and every other formula in the chapter is built from it.
2. When do you multiply and when do you add?
Multiply when stages happen together in sequence - shirt and trousers. Add when cases are mutually exclusive alternatives - a shirt or a jacket. The English words 'and' and 'or' are the signal, and misreading them is the most common structural error in this chapter.
3. Evaluate 5!.
120. 5! = 5 × 4 × 3 × 2 × 1 = 120. It counts the number of ways to arrange five distinct objects in a row. Worth memorising up to 6! = 720, since these values appear constantly and recomputing them wastes exam time.
4. How many three-digit numbers can be formed from the digits 1 to 9 without repetition?
504. There are 9 choices for the first digit, 8 for the second and 7 for the third, giving 9 × 8 × 7 = 504. This is ⁹P₃, and note that zero was excluded from the digit set - if 0 were allowed the first digit would need separate treatment, which is a standard complication.
5. How many three-digit numbers can be formed from the digits 1 to 9 with repetition allowed?
729. Each of the three positions can be filled in 9 ways independently, giving 9³ = 729. Compare with question 4: repetition being allowed means the pool does not shrink, which is why the answer is a power rather than a falling product.
6. Why is 0! defined as 1?
Because it makes the formulas consistent. ⁿCn should equal 1 - there is exactly one way to choose all n objects - and the formula n!/(n! 0!) gives that only if 0! = 1. It is a definition chosen so the general results hold without exceptions, not a computed value.
Practice this now
Practice set 2: permutations
7. Evaluate ⁷P₃.
210. ⁿPr = n!/(n − r)! = 7!/4! = 7 × 6 × 5 = 210. In practice, just multiply r descending factors starting from n - there is no need to compute the full factorials and divide. This counts ordered selections of 3 from 7.
8. In how many ways can n distinct objects be arranged in a row?
n!. The first position has n choices, the second n − 1, and so on down to 1. Every other permutation formula is a restriction of this one - ⁿPr arranges only r of them, and circular arrangements remove the rotational duplicates.
9. How many distinct arrangements are there of the letters of the word LEVEL?
30. There are 5 letters, but L appears twice and E appears twice. So the count is 5!/(2! × 2!) = 120/4 = 30. Using 5! = 120 would count each arrangement four times, because swapping the two Ls or the two Es produces no visible change.
10. In how many ways can 5 people be seated around a circular table?
24. Circular arrangements of n objects number (n − 1)!, so 4! = 24. Fixing one person's position removes the rotational duplicates - rotating everyone by one seat gives the same arrangement of neighbours. Answering 5! = 120 counts each seating five times.
11. In how many ways can 5 beads be strung on a necklace?
12. A necklace can be flipped over, so an arrangement and its mirror image are the same. The count is therefore (n − 1)!/2 = 4!/2 = 12. The distinction from question 10 is exactly whether the arrangement can be turned over - a table cannot, a necklace can.
12. In how many ways can 5 people be arranged in a row if two particular people must sit together?
48. Treat the pair as a single block, giving 4 units to arrange in 4! = 24 ways, then multiply by 2! = 2 for the two orders within the block: 24 × 2 = 48. The block method handles every 'must be together' constraint, and the complement gives the 'must not be together' count as 120 − 48 = 72.
Practice set 3: combinations
13. Evaluate ⁷C₃.
35. ⁿCr = n!/(r!(n − r)!) = 7!/(3! 4!) = 5040/144 = 35. Faster in practice: (7 × 6 × 5)/(3 × 2 × 1) = 210/6 = 35. Compare with question 7 - the same numbers give ⁷P₃ = 210, exactly 3! = 6 times larger, because each selection can be ordered in six ways.
14. Show that ¹⁰C₃ = ¹⁰C₇.
Both equal 120. The identity ⁿCr = ⁿC(n − r) holds because choosing which 3 to include is the same as choosing which 7 to exclude. This is useful computationally - always compute the version with the smaller r, since ¹⁰C₃ takes three multiplications and ¹⁰C₇ appears to take seven.
15. In how many ways can a committee of 3 be formed from 7 people?
35, which is ⁷C₃. Order does not matter here - a committee containing A, B and C is the same committee however it is listed. If the three posts were distinct, such as president, secretary and treasurer, the answer would instead be ⁷P₃ = 210.
16. From 5 men and 4 women, how many committees of 3 contain exactly 2 men?
40. Choose 2 men from 5 in ⁵C₂ = 10 ways and 1 woman from 4 in ⁴C₁ = 4 ways. The two choices happen together, so multiply: 10 × 4 = 40. Constrained-selection questions always factor this way - split by category, count each, then multiply.
17. How should you handle a question asking for 'at least one' of something?
Count the complement and subtract from the total. 'At least one' spans many cases while its complement - none - is a single case. Adding the cases individually is slower and risks omitting one, which is precisely the error the phrasing is designed to provoke.
18. What is the sum ⁿC₀ + ⁿC₁ + … + ⁿCn?
2ⁿ. It counts every possible subset of an n-element set, choosing 0, 1, 2 up to n elements. Equivalently, each element is independently either in or out, giving 2 × 2 × … = 2ⁿ. It also follows from the binomial theorem with both variables set to 1.
19. How many handshakes occur if 10 people each shake hands once with everyone else?
45. Each handshake involves an unordered pair, so the count is ¹⁰C₂ = (10 × 9)/2 = 45. Answering 10 × 9 = 90 counts each handshake twice, once from each participant's perspective - which is exactly the difference between a permutation and a combination.
Practice this now
Practice set 4: distribution
20. In how many ways can 10 identical balls be placed into 3 distinct boxes?
66. This is the stars-and-bars formula ⁿ⁺ʳ⁻¹C(r−1) with n = 10 and r = 3, giving ¹²C₂ = 66. The key feature is that the balls are identical, so only how many go into each box matters, not which ones. Boxes may be left empty here.
21. Same question, but each box must contain at least one ball.
36. Place one ball in each box first, leaving 7 to distribute freely - or use the formula ⁿ⁻¹C(r−1) = ⁹C₂ = 36. The 'at least one' condition is handled by pre-allocating the minimum and then applying the unrestricted formula to what remains, which is a pattern worth reusing.
22. How does the answer change if the 10 balls are distinct rather than identical?
It becomes 3¹⁰, since each ball independently chooses one of three boxes. That is 59,049 - vastly more than 66. Whether the objects are identical or distinct changes the problem completely, and it is the first thing to establish when reading a distribution question.
23. In how many ways can you select any number of items from 5 distinct items?
32, which is 2⁵. Each item is independently either selected or not. If the empty selection is excluded, the answer is 2⁵ − 1 = 31. Questions frequently specify 'at least one item', which is exactly that subtraction.
24. How many subsets does a set with 5 elements have?
32. This is the same count as question 23 - a subset is exactly a selection of elements, with each element either included or excluded. The number of proper subsets, excluding the set itself, is 31, and excluding the empty set as well gives 30.
Practice set 5: harder counting
25. From a group of 5 people, in how many ways can at least one be selected?
31. The total number of selections including the empty one is 2⁵ = 32, and subtracting the single empty case gives 31. This is the complement method in its simplest form, and it generalises directly to harder 'at least' conditions.
26. In how many arrangements of the letters of DAUGHTER do all the vowels come together?
4320. DAUGHTER has 8 distinct letters, of which A, U and E are vowels. Treat the three vowels as one block, giving 6 units to arrange in 6! = 720 ways, then arrange the vowels within the block in 3! = 6 ways: 720 × 6 = 4320. The block method from question 12 applies unchanged.
27. How many diagonals does a hexagon have?
9. The number of line segments joining pairs of vertices is ⁶C₂ = 15, but 6 of those are the sides of the polygon rather than diagonals. So the count is 15 − 6 = 9. The general formula ⁿC₂ − n follows from the same reasoning for any polygon.
28. What is meant by the rank of a word?
Its position when all arrangements of its letters are listed in dictionary order. It is computed by counting how many arrangements begin with a letter earlier in the alphabet, then how many share the first letter but have an earlier second letter, and so on - adding the counts and finally adding one. The technique is systematic counting rather than a formula.
29. What does the inclusion-exclusion principle state for two sets?
|A ∪ B| = |A| + |B| − |A ∩ B|. Adding the two counts includes the overlap twice, so it must be subtracted once. For three sets the pattern continues with the triple intersection added back. This is the counting analogue of the addition rule in probability.
30. How many shortest paths are there from one corner of a 3 × 2 grid to the opposite corner?
10. Any shortest path consists of 3 steps right and 2 steps up in some order, so the count is the number of arrangements of RRRUU, which is 5!/(3! 2!) = 10. Equivalently it is ⁵C₂ - choosing which 2 of the 5 steps are upward. Grid-path questions are always this repeated-arrangement problem in disguise.
How to study this chapter efficiently
- ✓For every question, write down two things before any formula: does order matter, and are the objects identical or distinct.
- ✓Learn the block method for 'together' constraints and the complement method for 'at least one'. Between them they cover a large share of the harder questions.
- ✓Practise varied situations rather than harder arithmetic. The computation is never the obstacle.
- ✓When a circular arrangement appears, ask whether it can be flipped. A table cannot, a necklace or garland can, and that halves the count.
- ✓Check answers for plausibility. A count that exceeds the total number of arrangements, or that is not a whole number, signals an error immediately.
- ✓Do this chapter before Probability. Most probability questions are counting problems with a division at the end.
Turn this into active practice
This is the chapter where reading solutions helps least. The modelling decision - permutation or combination, identical or distinct, together or apart - has already been made for you in a worked example, and that decision is the entire skill being tested.
The JEE Permutations and Combinations quiz on QUFF generates fresh questions across the counting principle, arrangements, selections, circular cases and distributions, marks them instantly and explains each answer. Do a mixed set and, for each error, write one line naming which modelling decision you got wrong. Those lines will repeat, and the repetition is the diagnosis.