The "art of counting" — arranging books on a shelf, forming a cricket team, or picking cards. The one question that decides the formula: does the order matter? This lesson keeps your pasted content but adds the why, worked placement examples, and the traps that cost marks.
| Rule | When | Operation | Example |
|---|---|---|---|
| AND Rule | Task A and Task B (both must happen) | Multiply | 3 shirts AND 2 pants = 3 × 2 = 6 outfits |
| OR Rule | Task A or Task B (either, not both) | Add | Bus (2 routes) OR Train (3 routes) = 2 + 3 = 5 ways |
Intuition: AND builds a tree — each shirt branches to 2 pants (3 branches × 2 leaves). OR builds parallel paths — you take one road or the other, so you add. Every P&C problem is a sequence of AND steps, sometimes with an OR fork. If you can phrase the task as "first do X, then do Y", multiply. If "either X or Y", add.
Placement cue: "and" in a sentence → usually AND. "or", "either", "cases" → OR. When a question has cases (e.g., at least 1 girl = 1 girl OR 2 girls OR 3 girls), you will add the cases — and each case internally is an AND of selections.
n! = n × (n−1) × (n−2) × … × 1 — the number of ways to order
n distinct items in a line.
5! = 5 × 4 × 3 × 2 × 1 = 1201! = 1, 0! = 1 — not a trick.
Why is 0! = 1? From combinations:
nCn = n! / (n! × 0!) = 1 for any n. The only
value that makes that true is 0! = 1. Also, there is exactly 1 way to arrange
nothing. Memorize factorials to 7 for speed:
| n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| n! | 1 | 1 | 2 | 6 | 24 | 120 | 720 | 5040 |
Speed note: 6! = 720 is your workhorse (appears in most P&C
cancellations). Keep 5! = 120 and 7! = 5040 on instant recall.
Use permutation when order matters — arranging books, seating people, forming numbers, assigning distinct roles like President / VP / Secretary.
nPr = n! / (n − r)! — n: total available,
r: number to arrange.
Derivation (why divide?): To pick and order r from
n, first position has n choices, second n−1, …, r-th
has n−r+1 choices. Multiply: n × (n−1) × … × (n−r+1) = n! / (n−r)!.
You are literally writing out n! and cancelling the unused tail.
Quick check: "Does swapping two chosen people give a different outcome?" — Yes → P. No (same team either way) → C.
n distinct items in a row:
n! (this is nPn).
n! / (p! × q! × …) where p, q are repeat counts.5! / 2! = 60. Deeper: MISSISSIPPI (11 letters: I×4, S×4, P×2) →
11! / (4! × 4! × 2!) = 34650. Each division collapses the identical swaps you
overcounted.
(n − 1)!n!? In a circle, rotating everyone one seat is the same
arrangement — you fix one person as reference and arrange the other
n−1 linearly. If clockwise ≠ anticlockwise (necklace that can be flipped),
divide by 2: (n−1)! / 2 — rarely asked, but know it.
5P4 = 5! / 1! = 120. If repetition were allowed,
it would be 54 = 625 (each of 4 places has 5 independent choices — AND
rule, not P). The formula changes completely when "without repetition" is removed.
Use combination when order does NOT matter — selecting a team, picking fruits, choosing cards, forming committees.
nCr = n! / [ r! × (n − r)! ]
Relationship:
nCr = nPr / r!
— you first count ordered picks (P), then divide by
r! because the r! orderings of the same selected set are
identical.
nC0 = 1 (one way to take nothing),
nCn = 1 (one way to take all)
nCr = nCn−r
— choosing 2 to keep = choosing 8 to reject from 10.
Placement trick: when r > n/2, compute the complement.
100C98 = 100C2 = 4950
in one step instead of huge factorials.
nCr + nCr−1 = n+1Cr
(Pascal's identity) — useful for telescoping sums, occasionally traps "at least" questions.
10C3 = 120. If the committee must have
at least 1 woman — see Method 3 below — do NOT use nP;
swapping two committee members is the same committee, so C is correct. If one of the three
seats were "chairperson", you'd switch to P for that seat.
Question type: arrange so that certain items are always together (e.g., vowels together, two people must sit together).
5! = 120 ways.5! = 120 ways.5! × 5! = 14400. If vowels were not all distinct (e.g., "APPLE" vowels
A,E), internal would be smaller.Question type: arrange so that certain items are never together (e.g., no two girls sit together, no two vowels adjacent).
m boys create m+1 gaps.
m+1Pk if order among restricted matters.
5! = 120.6P3 = 6×5×4 = 120 (equivalently
6C3 × 3!).5! × 6P3 = 120 × 120 = 14400.m+1 — if it does, answer is 0.
Question type: select at least 1, at most 2, etc.
At least 1 = Total − 0.
4C1×6C2=4×15=60, plus 4C2×6C1=6×6=36, plus
4C3=4 → 100.10C3 − 6C3 = 120 − 20 = 100
— one subtraction beats three cases.
Question type: distribute n identical items among
r distinct groups/bins.
n+r−1Cr−1 — think of n stars (★) and
r−1 bars (|) to split them. Arranging these n+r−1 symbols gives the
count.
n=10, r=3 → 10+3−1C3−1 = 12C2 = 667+3−1C2 = 9C2 = 36. This shift —
subtract mandatory 1s — is the most common stars-and-bars trap.rn (each item chooses a bin).
n people (each pair shakes once) =
nC2 — every unordered pair.
n-sided polygon =
nC2 − n — all connections between vertices minus the
n sides. Equivalently n(n−3)/2.
8C2 = 28.8C2 − 8 = 28 − 8 = 20 (or 8×5/2=20).| Ask | If Yes | If No |
|---|---|---|
| Does swapping two chosen items create a different result? | Permutation (nPr) | Combination (nCr) |
| Are you assigning distinct roles (captain, vice-captain)? | P — roles make order matter | C — just a team |
| Are you forming a number/word/order? | P | — |
| Are items identical? | Divide by repetition factorials; or stars and bars | Use factorials directly |
Common confusion: "Arrange 5 books on a shelf" vs "Choose 5 books to carry" —
same numbers, opposite formulas (5! vs nC5). The verb "arrange / order /
line up" → P. "choose / select / team / committee" → C. When a question mixes both (e.g.,
choose a committee then assign a president from it →
nCr × r), treat it as two AND steps.
| Symbol | Meaning |
|---|---|
n |
Total number of items |
r |
Number chosen or arranged |
! |
Factorial operator |
P |
Permutation (arrangement, order matters) |
C |
Combination (selection, order irrelevant) |
| Formula | When |
|---|---|
nPr = n!/(n−r)!
|
Arrange r from n (order matters) |
nCr = n!/[r!(n−r)!]
|
Choose r from n (order irrelevant) |
n!/(p!q!…) |
Permutation with repeats |
(n−1)! |
Circular arrangement |
n+r−1Cr−1
|
Identical items to distinct bins (stars and bars) |
nC2 /
nC2−n
|
Handshakes / Diagonals |
3 shirts AND 2 pants — total outfits (AND rule)?
What is 0! ?
When does order matter? Use?
APPLE (P repeated twice) — distinct arrangements?
Circular arrangement of n distinct people?
ⁿCr = ?
Vowels always together — which method?
No two girls together — which method?
n identical items among r distinct groups?
Diagonals in an n-sided polygon?
10 identical candies to 3 children (zero allowed) — ways?
5 boys, 3 girls, no two girls together — total?
At least 1 woman in committee of 3 from 6M+4W — fastest?
Order matters? ⇒ permutation ; not matters → combination
AND ⇒ multiply (eg 3 shirts AND 2 pants → 6) ; OR ⇒ add (eg 2 buses OR 3 trains → 5)
Decision ⇒ swap two chosen → different? Yes→P, No→C
n! ⇒ n×(n−1)×…×1 ; 5!=120, 6!=720, 7!=5040 ; 0! ⇒ 1 (makes ⁿCn=1)
ⁿPr ⇒ n!/(n−r)! ; n distinct in row ⇒ n!
Repeated ⇒ n!/(p!×q!…) (eg APPLE 5!/2!=60 ; MISSISSIPPI 11!/(4!4!2!))
Circle ⇒ (n−1)! ; necklace (flip same) ⇒ (n−1)!/2
ⁿCr ⇒ n!/[r!×(n−r)!] ; = ⁿPr/r!
ⁿC0=1 ; ⁿCn=1 ; ⁿCr=ⁿCn−r (use when r>n/2)
a) Bundle/string ⇒ together → bundle + internal (eg vowels together: 5 units→5! × internal 5!)
b) Gap ⇒ separated → arrange unrestricted → gaps (m+1) → ⁽ᵐ⁺¹⁾Pk
(eg 5 boys, 3 girls→5!×⁶P₃=14400)
c) At least/at most ⇒ direct sum OR shortcut total − forbidden
(eg ≥1W: ¹⁰C₃−⁶C₃=100 faster than 3 cases)
d) Stars and bars ⇒ n identical among r groups → ⁿ⁺ʳ⁻¹Cr−1
Zero allowed: ¹²C₂=66 ; ≥1 each: ⁹C₂=36
e) Handshakes ⇒ ⁿC₂ ; Diagonals ⇒ ⁿC₂−n = n(n−3)/2
Bundle×internal × ; gap count = m+1 ; circular (n−1)! ; repeats divide ; ≥1 → complement
n ⇒ total ; r ⇒ chosen/arranged ; ! ⇒ factorial ; P/C ⇒ perm/comb
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