Back to Topic 1.10 — Counting & binomial (HL only)
1.10.1Math AA HL10 flashcards

Arrangements (order matters)

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Card 1 of 101.10.1
1.10.1
Question

What makes a counting question an 'arrangement'?

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All 10 Flashcards — Arrangements (order matters)

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Card 1concept

Question

What makes a counting question an 'arrangement'?

Answer

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

Card 2concept

Question

Describe the box method for arrangements.

Answer

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

Card 3concept

Question

How do you count r-digit numbers with no leading zero?

Answer

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

Card 4formula

Question

How do you arrange ALL n distinct objects in a row?

Answer

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

Card 5formula

Question

State the formula for ⁿPᵣ and what it counts.

Answer

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

Card 6concept

Question

How is ⁿPᵣ just the multiplication idea?

Answer

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

Card 7concept

Question

Arrange ALL of them vs SOME of them — which formula?

Answer

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

Card 8concept

Question

Seats, finishing orders, codes — arrangement or not?

Answer

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

Card 9concept

Question

On Paper 2, where is nPr on the TI-84?

Answer

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

Card 10formula

Question

What is ⁿPₙ equal to?

Answer

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

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