Permutations and Combinations
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Direct answer
Every counting problem is a chain of two decisions — selecting objects and arranging them — governed by the multiplication and addition principles. Selections use combinations C(n, r) = n!/(r! (n - r)!) when order does not matter, arrangements use permutations P(n, r) = n!/(n - r)! when it does, and repetition, identical objects and circular seating each modify these through fixed adjustments. Spotting which features a question contains is most of the work.
What you must remember
- Fundamental principles: tasks performed together multiply (m × n ways), tasks as alternatives add (m + n ways); apply them slot by slot when objects are placed with conditions.
- P(n, r) = n!/(n - r)! for ordered arrangements without repetition; C(n, r) = n!/(r! (n - r)!) for unordered selections; C(n, r) = C(n, n - r) and C(n, r) + C(n, r - 1) = C(n + 1, r).
- Words with repeated letters: n!/(p! q! ...) where p, q are the repeat counts; arranging all distinct objects gives plain n!.
- Circular arrangements: (n - 1)! ways around a round table, because one person's position fixes the frame; necklaces and garlands give (n - 1)!/2 since clockwise and anticlockwise orders coincide.
- Selection counts: n distinct objects give 2^n - 1 non-empty selections; "at least one" is usually fastest through the complement.
- Stars and bars: the non-negative integer solutions of x1 + x2 + ... + xr = n number C(n + r - 1, r - 1); positive solutions need C(n - 1, r - 1).
- Divisors: N = p^a q^b ... has (a + 1)(b + 1) ... divisors; gap method: arrange the unrestricted group first, then place "no-two-together" objects in the gaps.
Common confusion
The trap is order blindness — treating a committee as an arrangement or a queue as a selection. Ask "would swapping two chosen objects create a new outcome?"; if yes, order matters. The identical-objects filter is the second killer: two arrangements that differ only by swapping identical letters are the same arrangement, hence the division by the repeat factorials. And whenever a question says "at least", check whether counting the complement is cheaper before building cases.
Exam-focused takeaway
JEE Main asks word arrangements, digit numbers under conditions, committee selections, circular seating and divisor counts — method recognition converts these into one-line numericals. JEE Advanced layers the same machinery: distributions with group sizes, inclusion-exclusion with overlapping conditions, and counting that later feeds probability questions. Draw the slots, mark each restriction, multiply deliberately; most wrong answers are misreadings, not hard mathematics.
Frequently asked questions
When do I use C(n, r) instead of P(n, r)?
Whenever order among the chosen objects does not create a new outcome — committees, hands of cards, selected questions. If the arrangement of the chosen set matters, use P(n, r).
How many ways can n people sit around a round table?
(n - 1)!, because rotations of the same seating are identical; if reflections also count as the same, as in a necklace, halve it.
How are repeated letters handled in word arrangements?
Arrange all letters and divide by the factorials of the repeat counts; for example, arrangements of ALLAHABAD-style words follow n!/(p! q! ...).
What is the stars and bars method for?
Counting solutions of x1 + x2 + ... + xr = n in non-negative integers — distributing n identical items into r distinct boxes — giving C(n + r - 1, r - 1).
How do I arrange people so that two particular ones never sit together?
Arrange the rest and place the pair in the gaps; total minus together works equally well.