← Maths Extension 1 syllabus

NSWMaths Extension 1

Combinatorics (ME-A1)

7 dot points across 7 inquiry questions. Click any dot point for a focused answer with worked past exam questions where available.

When objects are arranged in a closed ring rather than a line, which arrangements should count as the same, and how does that change the count from n! to (n-1)!?

Count arrangements of n distinct objects around a circle as (n-1)! by treating rotations as identical, extend this to groups and blocks around a circle, alternating patterns and 'not together' restrictions handled by the complement, and recognise that for a necklace or bracelet, where a reflection also coincides, the count is (n-1)!/2
Arranging objects around a circle. Why n distinct objects give (n-1)! when rotations count as the same seating, the block method for groups and couples (2^k x (k-1)!), alternating boys and girls, 'not together' by the complement, and necklaces/bracelets where reflections also coincide so you divide by 2.
18 min answer →

When some of the objects being arranged are identical, why does plain n! overcount, and how do you divide out the repeats to count only the genuinely different arrangements?

Count the distinct arrangements of n objects when some are identical: divide n! by the factorial of each repeat count to get n! over r1! r2! ... rk!, and handle the two-type special case where every object is one of two kinds (2^n arrangements in all, or choose the positions of one kind with a combination)
Why arranging objects with repeats needs n! over r1! r2! ... rk!: plain n! overcounts because swapping identical objects changes nothing, so you divide by the factorial of each repeat count. Covers word arrangements like PRESSES, the binary two-type case (2^n or choose positions with a combination), cases, and separating identical items.
18 min answer →

When the order of a selection does not matter, how do you count the choices, and how do the restriction moves you learned for arrangements carry across to unordered selections?

Count unordered selections (combinations) using ^{n}C_{r} = ^{n}P_{r}/r! = n!/(r!(n-r)!), recognise that an n-element set has 2^n subsets and that the r-element subsets number ^{n}C_{r} with the symmetry ^{n}C_{r} = ^{n}C_{n-r} and row sum 2^n, and apply the at-least, complement, conditional and geometric counting techniques to selection problems
Counting unordered selections with combinations. Why nCr = nPr/r! = n!/(r!(n-r)!), why an n-set has 2^n subsets, the symmetry nCr = nC(n-r), and the row sum 2^n. Restriction technique with combinations: at least k by complement, must contain or exclude an item, committees with constraints, and geometry counts.
18 min answer →

When every outcome is equally likely, how do you turn a probability into a ratio of two counts, and how do the counting moves (multiplication principle, permutations, combinations, complement, cases) supply both the favourable count and the total?

Compute probabilities of equally likely outcomes as P = (favourable outcomes)/(total outcomes) where both counts come from the counting methods (multiplication principle, ^{n}P_{r} and ^{n}C_{r}), and apply the complement, separate-pool multiplication, cases and inclusion-exclusion to find the favourable count for selection, card-hand, digit-string and with/without-replacement problems
Turning counting into probability. When outcomes are equally likely, P = favourable/total and both counts come from the multiplication principle, nPr and nCr. Committee and card-hand probabilities by combinations, exactly-k of a type, at-least-one by the complement, a five-digit-number probability, inclusion-exclusion, and with versus without replacement.
19 min answer →

What does the factorial n! count, and how do you simplify expressions built from factorials?

Define and evaluate factorials, use the recursive relation n! = n(n-1)!, and simplify factorial expressions and fractions by unrolling and cancelling
A first-contact answer to the HSC Maths Extension 1 dot point on factorial notation. What n! means, why 0! = 1, the recursive rule n! = n(n-1)!, and how unrolling cancels factorial fractions, combines them over a common factorial, and counts trailing zeroes, with the arranging-in-a-row motivation and worked examples.
12 min answer →

When an ordered arrangement carries a restriction, how do you choose between grouping items into a block, counting the complement, and splitting into cases with an overlap correction?

Apply three reusable counting principles to ordered arrangements with restrictions: keep tied items together by grouping them into a block and ordering inside it, count an unwanted condition and subtract it from the total (the complement), and split an 'or' count into non-overlapping cases, subtracting the overlap by inclusion-exclusion when the cases meet
The three restriction moves the HSC Extension 1 counting dot point keeps testing. Group tied items into a block and order inside it, count the unwanted and subtract it (the complement), and split an 'or' count into cases, subtracting the overlap by inclusion-exclusion. Includes a symmetry shortcut, slot diagrams and a Venn picture.
17 min answer →

How do you count the number of ordered ways to make a sequence of choices, and how does allowing or forbidding repetition change the count?

Use the multiplication principle to count ordered selections across stages, count ordered selections with repetition as n^r and without repetition as nPr = n!/(n-r)!, and apply restriction techniques: deal with the difficulty first, fix a position, keep items together as a block, and count 'at least one' by the complement
A first-contact answer to the HSC Maths Extension 1 dot point on counting ordered selections. The multiplication principle across stages, ordered selections with repetition (n^r) versus without (nPr), and the restriction moves: deal with the difficulty first, fix a position, keep items together as a block, and count at-least-one by the complement, using the filled-box slot method.
16 min answer →