Counting & binomial (HL only)

Order matters — the position of each object counts, so ABC and CBA are different. Count by filling positions and multiplying.

Draw one box per position, write how many choices go in each box, then multiply. Fill the most restricted box first.

The first box has 9 choices (1–9, not 0); fill the rest from the remaining digits, then multiply.

n! = n × (n − 1) × … × 1. Example: 9 people in a line = 9! = 362880.

ⁿPᵣ = n!/(n − r)! — the number of ways to arrange r objects out of n in a definite order.

It multiplies r numbers counting down from n: n × (n − 1) × … (r factors). e.g. ⁸P₃ = 8 × 7 × 6 = 336.

All n → n!. Just r of them, in order → ⁿPᵣ = n!/(n − r)!.

Arrangements — the position matters, so use n! (all) or ⁿPᵣ (some), filling positions and multiplying.

MATH, arrow right to PRB, then 2: nPr. Type n, choose nPr, type r, ENTER.

n! — arranging all n in order (since n!/(n − n)! = n!/0! = n!).

Does order matter? Arrange/rank/roles → yes (ⁿPᵣ); choose/team/committee → no (ⁿCᵣ). Then check for restrictions.

Block method — glue them into one item, arrange, then multiply by the internal orders (k!).

Total − together (for a pair), or the gap method (seat the others, drop the rest into separate gaps).

Fix that member in place, then choose the rest from those remaining.

Add up the allowed cases, or use total − unwanted (the complement).

Divide n! by the factorial of each repeated letter: n! ÷ (p! q! …).

Choose which of the moves go 'up' (or right): ⁿCᵣ.

Because the wording sets the type. Medals (distinct roles) → ⁿPᵣ; a committee (no roles) → ⁿCᵣ.

When n is negative or a fraction (not a positive whole number) — the terms never stop.

1 + nx + n(n−1)/2! x² + n(n−1)(n−2)/3! x³ + … , valid for |x| < 1.

Multiply by the next falling factor of n on top and divide by the next factorial: n, then n(n−1)/2!, then n(n−1)(n−2)/3!.

Replace every x in the formula with kx, then simplify — remember to square and cube the k.

1 − x + x².

1 + 6x + 27x².

n.

n(n − 1)/2.

T_(r+1) = ⁿCᵣ aⁿ⁻ʳ bʳ — choose r of the b's.

Use the general term; set the power of x equal to k to fix r, then compute ⁿCᵣ aⁿ⁻ʳ bʳ.

Write the coefficient as an expression in the unknown, set it equal to the given number, and solve.

ⁿC₂ a².

⁵C₃ × 2² = 10 × 4 = 40.

10a² = 40 ⇒ a² = 4 ⇒ a = 2.

Forgetting to raise the inside coefficient to the power r — (2x)² = 4x², not 2x².

The coefficient counts the ⁿCᵣ ways that term is formed — the powers alone aren't enough.

When |x| < 1 — only then do the terms shrink so the series settles to a value.

|kx| < 1, i.e. |x| < 1/|k|.

Choose x so the bracket equals the target number (x should be small), then evaluate the first few terms.

x = 0.02: 1 + 0.01 − 0.00005 = 1.00995.

Only then do the powers of x get smaller, so the later terms fade and the sum converges.

|2x| < 1, so |x| < ½.

x = 0.04 (since 1 + x = 1.04).

Yes — the later (ignored) terms are even tinier, so the first few terms are more accurate.

When you pick a GROUP and the order inside doesn't matter — a team, a committee, a sample. {A,B} is the same as {B,A}.

Top: r numbers counting down from n. Bottom: r! = r × (r − 1) × … × 1. e.g. ⁵C₂ = (5 × 4)/(2 × 1) = 10.

Picking r in order counts each group r! times (its r! orderings). A group has no order, so divide those out.

Ask: does the order matter? Order matters → arrangement (ⁿPᵣ). Order doesn't → selection (ⁿCᵣ).

ⁿCᵣ = n!/(r!(n − r)!) — the number of ways to choose r objects from n with order ignored.

(8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56.

MATH, arrow right to PRB, then 3: nCr. Type n, choose nCr, type r, ENTER.

ⁿCᵣ — these are unordered groups, so order doesn't matter.

Choose — the increasing order is fixed, so each set of digits gives exactly one number. Count with ⁿCᵣ.

Once you pick the items, there is only ONE way to put them in increasing order — nothing is left to arrange, so order doesn't add anything.

Choose 4 from 1–9 (0 can't appear): ⁹C₄ = 126.

'increasing order', 'decreasing order', 'alphabetical order' — all fix the order, so use ⁿCᵣ.

In increasing order 0 would have to come first, but a number can't start with 0. So choose from 1–9.

Choose 3 of the 7 letters: ⁷C₃ = 35 (alphabetical order is automatic).

You counted the orderings, but the order is forced (one per set). Divide out — i.e. use ⁿCᵣ instead.

No — both fix the order, so both are ⁿCᵣ. Each chosen set has one increasing and one decreasing arrangement.

Put that person in first (they use one place), then choose the rest from the people who are left.

Captain in, choose 3 from the remaining 9: ⁹C₃ = 84.

That person is unavailable, so just choose from everyone else.

Choose all 4 from the other 9: ⁹C₄ = 126.

An ordinary selection (ⁿCᵣ) on the remaining people for the remaining places.

Both drop by 1: one fewer place to fill AND one fewer person to choose from.

The number of people drops by 1, but the number of places stays the same.

'must include', 'always plays', 'the chair is on the committee', or 'must not be chosen / refuses'.

Choose from each group separately, then multiply (AND → ×). e.g. 2 men and 2 women = ⁵C₂ × ⁶C₂.

Each way of choosing the first group can pair with each way of choosing the second — that's the multiplication principle.

Add the cases: exactly 2 + exactly 3 + … Each case = choose that many women AND the rest from the other group.

When there are many 'at least' cases but few to exclude — e.g. 'at least 1' = total − (none).

The other 2 must be women, so multiply ⁵C₂ (men) × ⁶C₂ (women) — the numbers add to 4.

Exactly 0 girls + exactly 1 girl, added together.

Adding the two ⁿCᵣ values instead of multiplying them (it's AND, not OR).

Split the numbers into remainder groups (mod 3), then count the choices that make the total work — choosing from each group, multiplying and adding cases.

Glue them into one block, arrange the blocks, then multiply by the ways to arrange inside the block.

k! — the ways to arrange the k people inside the block.

Treat AB as one block. Arrange the (n − 1) things, then × 2 for the AB / BA orders inside.

'Together' allows both internal orders (× 2). 'Immediately after' fixes the order (× 1).

One fewer for each extra person in the block — a block of k among n total leaves (n − k + 1) things to arrange.

Glue the 3: arrange 5 things (5!) × 3! inside = 120 × 6 = 720.

The order inside the block is fixed (only one way), so there's nothing extra to multiply by.

'together', 'next to each other', 'side by side', 'consecutive', 'immediately after/before'.

Total − together. Count all arrangements, then subtract the ones where the two ARE together (block method).

Gap method: arrange the other items first; they make gaps; place the restricted items into different gaps.

n + 1 gaps — one between each pair plus one at each end.

They go into different gaps and the items are distinct, so the order in which gaps are filled matters → ⁿPᵣ.

5! − (4! × 2) = 120 − 48 = 72.

Boys 4! = 24, 5 gaps, girls ⁵P₂ = 20: 24 × 20 = 480.

A single pair not together → total − together. A whole group with none adjacent → the gap method.

'not next to', 'not adjacent', 'apart', 'at least one seat between', 'not sharing a boundary'.

A route is an arrangement of right and up moves. Choose which moves go 'up' (or right) with ⁿCᵣ.

Make a right moves and b up moves (a + b total); choose the b up moves: ⁽ᵃ⁺ᵇ⁾Cᵦ.

Divide n! by the factorial of each repeated letter: n! ÷ (p! q! …).

Swapping two identical letters gives the same word, so plain n! counts each arrangement several times; dividing removes the duplicates.

6 letters with A×3, N×2: 6! ÷ (3! 2!) = 720 ÷ 12 = 60.

7 moves, choose 3 up: ⁷C₃ = 35.

A grid route is a word made of two letters (R and U), so both use the same arrangement idea.

11 letters with S×4, I×4, P×2: 11! ÷ (4! 4! 2!) = 34650.

ⁿP₂ = n(n − 1). Set n(n − 1) = k, rearrange to a quadratic, solve, and keep the positive whole-number root.

ⁿC₂ = n(n − 1)/2. Multiply by 2 to get n(n − 1) = 2k, solve the quadratic, keep the positive root.

n counts objects, so it must be a positive whole number — a negative root has no meaning.

n(n − 1).

n(n − 1)/2.

a + b = n, because ⁿCᵣ = ⁿCₙ₋ᵣ. e.g. ⁿC₃ = ⁿC₇ ⇒ n = 10.

n(n − 1) = 42 ⇒ (n − 7)(n + 6) = 0 ⇒ n = 7.

n(n − 1) = 56 ⇒ (n − 8)(n + 7) = 0 ⇒ n = 8.