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Year 11 Specialist (Unit 1 & 2) Combinatorics

Permutations with restrictions

20 practice questions 0 video lessons Theory + worked examples

Master permutations with restrictions for Year 11 Specialist Mathematics in Queensland (QCAA). They are ordered arrangements of distinct objects where a rule, such as keeping certain objects together or apart, limits which arrangements count.

You will learn the block method for objects that must stay together, the gap method for keeping objects apart, and how to count alternating and fixed-position arrangements — key skills for probability and QCAA combinatorics.

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Theory

Permutations with restrictions are ordered arrangements of distinct objects where a condition limits which arrangements are allowed — a core skill of Year 11 Specialist Mathematics (QCAA, Queensland). This page shows the block method for keeping objects together, the gap method for keeping them apart, and how to handle alternating and fixed-position problems.

A permutation is an ordered arrangement of distinct objects. With no conditions, \(n\) distinct objects arrange in a row in \(n!\) ways. A restriction is a rule — such as "these two must be together" or "no two of these may touch" — that rules some of those arrangements out.

When certain objects must stay together, use the block method: tie those objects into a single unit (a "block"), arrange the units, then arrange the objects inside the block. Treating \(k\) objects as one block turns \(n\) objects into \(n-k+1\) units.

When objects must be kept apart (not next to each other), you have two routes. Subtract: count every arrangement, then take away the ones where they are together. Or use the gap method: seat the other objects first, then drop the restricted objects into the separate gaps between and around them, so no two can touch.

Two more common restrictions: alternating arrangements (two kinds of object must take turns), counted by fixing the pattern then arranging each kind; and fixed positions (an object is pinned to a particular seat or to an end), counted by locking that object in place and arranging the rest.

Block method diagram Three girls are boxed together as one unit, standing beside two separate boy boxes. That gives three units arranged in 3 factorial ways, and the three girls inside the block in 3 factorial ways, so the total is 3 factorial times 3 factorial = 36. girls as ONE unit (3 girls inside) boy boy unit 1 unit 2 unit 3 3 units × block inside = 3! × 3! = 36
Block method: \(3\) girls tied into one unit gives \(3\) units, so \(3!\times3!=36\).
Gap method diagram Four boy boxes are seated in a row. Between and around them are five dashed gap slots, numbered 1 to 5. Each girl drops into a separate gap so no two girls touch: seat four boys in 4 factorial ways, then place three girls in the five gaps in 5 times 4 times 3 ways, giving a total of 1440. 1 2 3 4 5 B1 B2 B3 B4 5 gaps (dashed) for the girls 4! × (5 × 4 × 3) = 1440
Gap method: seat \(4\) boys (\(4!\)), then place \(3\) girls in the \(5\) gaps: \(4!\times5\times4\times3=1440\).

Objects together (block method). Tie the \(k\) objects that must stay together into one block. The \(n\) objects become \(n-k+1\) units; arrange the units, then the block internally:

\[ N = (n-k+1)! \times k! \]
N=(nk+1)!×k!

Objects apart (subtraction). Take the together-count away from the total:

\[ N_{\text{apart}} = n! - (n-k+1)!\times k! \]
Napart=n!(nk+1)!×k!

Objects apart (gap method). Seat the \(m\) other objects (\(m!\) ways). They create \(m+1\) gaps; slot the \(k\) restricted objects into separate gaps, order mattering:

\[ N = m! \times \dfrac{(m+1)!}{(m+1-k)!} \]
N=m!×(m+1)!(m+1k)!

Alternating (two kinds, \(p\) of one and \(q\) of the other). Count the valid patterns (\(2\) when \(p=q\), otherwise \(1\)), then arrange each kind:

\[ N = (\text{patterns}) \times p! \times q! \]
N=patterns×p!×q!
The block still gets arranged inside. A block of \(k\) objects counts as one unit for placing, but the objects within it can still be ordered in \(k!\) ways — never forget the \(\times k!\).

How to count a restricted arrangement

  1. Name the restriction: are objects forced together, kept apart, made to alternate, or fixed to a position or end?
  2. Pick the technique: block method for together; subtraction or the gap method for apart; pattern-count for alternating; lock-then-arrange for a fixed position.
  3. Count each stage: the units (or seated objects and their gaps), and any internal arrangement of a block — watch that a block still needs \(\times k!\).
  4. Combine: multiply the stages together; for "apart" by subtraction, take the together-count away from the total.
Example 1 — Two together (block method)
Five people stand in a row. In how many arrangements do two particular people, Jack and Kim, stand next to each other?
Solution

Tie Jack and Kim into one block, then arrange the units:

\(\text{units (block + 3 others)}\)\(=\)\(4\)
\(\text{arrange the units}\)\(=\)\(4!\)
\(\text{Jack and Kim inside the block}\)\(=\)\(2!\)
\(\text{total}\)\(=\)\(4! \times 2!\)
\(=\)\(24 \times 2\)
\(=\)\(48\)

\(48\) arrangements.

Example 2 — Two apart (total minus together)
Five people stand in a row. In how many arrangements do Jack and Kim not stand next to each other?
Solution

Count every arrangement, then subtract the together ones from Example 1:

\(\text{all arrangements}\)\(=\)\(5!\)
\(=\)\(120\)
\(\text{together (block)}\)\(=\)\(4! \times 2!\)
\(=\)\(48\)
\(\text{apart}\)\(=\)\(120 - 48\)
\(=\)\(72\)

\(72\) arrangements.

Example 3 — None adjacent (gap method)
Four boys and three girls stand in a row so that no two girls are next to each other. In how many ways can this be done?
Solution

Seat the boys first, then drop the girls into the separate gaps between them:

\(\text{seat the 4 boys}\)\(=\)\(4!\)
\(=\)\(24\)
\(\text{gaps created (before, between, after)}\)\(=\)\(5\)
\(\text{place 3 girls in the gaps (order matters)}\)\(=\)\(5 \times 4 \times 3\)
\(=\)\(60\)
\(\text{total}\)\(=\)\(24 \times 60\)
\(=\)\(1440\)

\(1440\) arrangements.

Example 4 — Alternating
Three boys and three girls stand in a row so that the genders alternate (no two boys and no two girls next to each other). In how many ways can this be done?
Solution

Count the patterns first, then arrange each gender within its own places:

\(\text{patterns (BGBGBG, GBGBGB)}\)\(=\)\(2\)
\(\text{arrange the boys}\)\(=\)\(3!\)
\(\text{arrange the girls}\)\(=\)\(3!\)
\(\text{total}\)\(=\)\(2 \times 3! \times 3!\)
\(=\)\(2 \times 6 \times 6\)
\(=\)\(72\)

\(72\) arrangements.

Common pitfalls

Forgetting to arrange inside the block. Watch out for stopping at \((n-k+1)!\). A block of \(k\) objects still has \(k!\) internal orders — always multiply by \(k!\).
Counting the block as more than one unit. Once objects are tied together they occupy a single position in the row of units, so \(n\) objects become \(n-k+1\) units, not \(n\).
Using the wrong number of gaps. Seating \(m\) objects creates \(m+1\) gaps (one at each end plus the ones between), not \(m\). And the gaps are ordered, so placing objects into them is a permutation.
Adding when a fixed object should just be locked. If a person is fixed at one end, that end is settled in \(1\) way; simply arrange the remaining \(n-1\) objects in \((n-1)!\) ways.

Frequently asked questions

What is the block method for permutations?

When objects must stay together, tie them into a single block and treat it as one unit. Arrange the units, then arrange the objects inside the block. For \(k\) objects together among \(n\), that is \((n-k+1)!\times k!\).

How do you arrange objects so that two are not together?

Either subtract the together-count from the total number of arrangements, or use the gap method: seat the other objects first, then place the restricted objects into separate gaps so they cannot touch.

What is the gap method?

Seat the unrestricted objects first. Between and around them there are gaps — one more gap than the number seated. Drop the restricted objects into separate gaps, and because order matters this is a permutation of the gaps.

How many ways can boys and girls alternate in a row?

Count the valid alternating patterns first (two patterns when the numbers are equal, one otherwise), then multiply by the arrangements of each group. For three boys and three girls it is \(2\times3!\times3!=72\).

How do you handle a person fixed at one end of a row?

Lock that person into the fixed place, which happens in \(1\) way, then arrange everyone else. For six people with one fixed at the left end that is \(5!=120\) arrangements.

Why do you multiply by k! inside a block?

The block fixes which objects sit together, but not their order within the group. Those \(k\) objects can still be arranged among themselves in \(k!\) ways, so every unit-arrangement splits into \(k!\) real arrangements.