The Chapter in One Idea

Everything here flows from one principle:

Successive choices multiply. "mm ways, following which nn ways" gives m×nm \times n ways. Arrangements (nPr^nP_r), repetition counts (nrn^r), repeated-letter words and selections (nCr^nC_r) are all this principle plus one correction — divide out whatever you have overcounted.

Before any formula, ask two questions: does order matter? and is repetition allowed? The answers pick the tool; the rest is arithmetic.

Counting and Arrangement — Formula Card

  • Multiplication principle: successive events with mm, then nn ways: m×nm \times n (extends to any number of stages). Add across mutually exclusive cases instead.
  • Factorial: n!=1×2×⋯×nn! = 1 \times 2 \times \cdots \times n; 0!=10! = 1; n!=n (n−1)!n! = n\,(n-1)!. Cancel tails: 7!5!=42\frac{7!}{5!} = 42; factor out: 7!−5!=41×5!7! - 5! = 41 \times 5!.
  • Permutations (order matters, no repetition): nPr=n!(n−r)!=n(n−1)⋯(n−r+1)^nP_r = \frac{n!}{(n-r)!} = n(n-1)\cdots(n-r+1); nPn=n!^nP_n = n!.
  • With repetition allowed: nrn^r.
  • Repeated objects: nn objects with kinds repeated p1,p2,…,pkp_1, p_2, \ldots, p_k times arrange in n!p1! p2!⋯pk!\frac{n!}{p_1!\,p_2!\cdots p_k!} ways. Write the letter inventory first.
  • The zero trap: strings with a leading 0 are not numbers — lock the first place to nonzero, or subtract the leading-zero strings.
  • Vacant-places habit: fill the most restricted place first (even units digit, fixed first letter).

Combinations — Formula Card

  • Selections (order does not matter): nCr=n!r! (n−r)!^nC_r = \frac{n!}{r!\,(n-r)!}; nC0=nCn=1^nC_0 = {^nC_n} = 1; nC1=n^nC_1 = n.
  • The bridge: nPr=nCr×r!^nP_r = {^nC_r} \times r! — select, then arrange.
  • Symmetry: nCr=nCn−r^nC_r = {^nC_{n-r}} — compute 50C48^{50}C_{48} as 50C2^{50}C_2.
  • Equality rule: nCa=nCb⇒a=b^nC_a = {^nC_b} \Rightarrow a = b or a+b=na + b = n.
  • Pascal identity: nCr+nCr−1=n+1Cr^nC_r + {^nC_{r-1}} = {^{n+1}C_r}.
  • Multi-group selections multiply: 3 boys from 5 AND 3 girls from 4: 5C3×4C3^5C_3 \times {^4C_3}.
  • At-least / at-most: split into exact cases and add — or subtract the complement from the total. Check every case against the available stock.
  • All subsets: nC0+nC1+⋯+nCn=2n^nC_0 + {^nC_1} + \cdots + {^nC_n} = 2^n; non-empty selections from nn distinct objects: 2n−12^n - 1; from identical stock, counts multiply as (p1+1)(p2+1)⋯−1(p_1 + 1)(p_2 + 1)\cdots - 1.

JEE Quick Card — Restrictions and Circles

  • Bundle (together): glue the group, arrange the units, multiply by internal arrangements: DAUGHTER vowels together =6!×3!=4320= 6! \times 3! = 4320.
  • Complement (not all together): total −- together: 8!−4320=360008! - 4320 = 36000.
  • Gap method (no two adjacent): arrange the free objects, place the restricted ones in distinct gaps: 5 girls, 3 boys =5!×6P3=14400= 5! \times {^6P_3} = 14400. Identical objects pick gaps by nCr^nC_r, distinct ones by nPr^nP_r.
  • Circular permutations: nn distinct objects in a circle: (n−1)!(n-1)!; necklace/garland (flips identical): (n−1)!2\frac{(n-1)!}{2}; numbered seats restore n!n!.
  • Dictionary rank: count complete alphabetical blocks before the word, then walk the final block; repeated letters divide block counts.
  • Fixed positions: pin the fixed letters/digits, arrange the rest (INDEPENDENCE starting with P: 11!3! 4! 2!\frac{11!}{3!\,4!\,2!}).
  • Sum of all numbers formed: each digit sits in each place (total)(digits)\frac{(\text{total})}{(\text{digits})} times; multiply by the repunit (111…111\ldots).

Last-Minute Mistake Checklist

Before the exam, scan this list — each item is a real mark lost by real students:

  1. Using nPr^nP_r for committees or nCr^nC_r for words — decide order matters or not before touching a formula.
  2. Multiplying when cases are alternatives: "2 or 3 or 4 flags" adds; "pant then shirt" multiplies.
  3. Forgetting the repetition clause: no-repetition shrinks choices box by box; with repetition every box stays full (nrn^r).
  4. Counting leading-zero strings as numbers — subtract them or lock the first digit.
  5. Missing a repeated letter in the inventory (MISSISSIPPI has 4 I's AND 4 S's AND 2 P's).
  6. Bundling for "no two adjacent" — that needs the gap method; the bundle-complement only handles "not all together".
  7. Ordering identical objects in gaps: identical I's choose gaps by 8C4^8C_4, not 8P4^8P_4.
  8. Circular counts: forgetting to fix an anchor ((n−1)!(n-1)!, not n!n!) — or applying (n−1)!(n-1)! when the seats are numbered.
  9. Dropping the second root of nCa=nCb^nC_a = {^nC_b} (a+b=na + b = n) — or keeping an algebraic root that violates r≤nr \le n.
  10. "At least one" the long way — complement (total minus none) is one line; case-adding invites arithmetic slips.