Arrangements (Permutations) Revision Notes for Leaving Cert Mathematics

Arrangements (Permutations)

What are arrangements (permutations)?

An arrangement or permutation refers to the different ways we can order or sequence a collection of distinct objects. When we change the order of objects, we create a new arrangement.

For example, if we have three letters A, B, and C, we can arrange them in these different ways:

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All possible arrangements of A, B, C:

ABC, ACB, BAC, BCA, CAB, CBA

This gives us six different arrangements in total.

The box method

A helpful way to think about arrangements is using the box method. Imagine we have empty boxes representing positions, and we need to fill them with our objects.

For three objects, we think about how many choices we have for each position:

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Box Method Logic:

First box: 3 choices (any of the three objects)
Second box: 2 choices (any of the remaining two objects)
Third box: 1 choice (the last remaining object)

Therefore: $3 \times 2 \times 1 = 6$ different arrangements.

The factorial formula

Definition of factorial

We use factorial notation to represent these calculations. The symbol $n!$ (read as "n factorial") means:

$n! = n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1$

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Factorial Examples:

$3! = 3 \times 2 \times 1 = 6$
$4! = 4 \times 3 \times 2 \times 1 = 24$
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

General rule for permutations

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Fundamental Permutation Rule:

The number of arrangements (permutations) of n different objects is $n!$

Worked example 1: Basic permutation

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Worked Example: Six-Digit Numbers

Question: How many different six-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6 using all the digits in each number?

Solution: The six digits can be arranged in $6!$ ways. $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

Answer: 720 different numbers can be formed.

Permutations with restrictions

When we have restrictions (conditions that limit where certain objects can be placed), we use a modified approach:

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Key Strategy: Fill restriction boxes first

Identify the restriction and fill those positions first
Count the remaining objects and positions
Fill the remaining boxes using the box method

Worked example 2: Single restriction

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Worked Example: TUESDAY Arrangements

Question: The letters of the word TUESDAY are arranged in a line. (i) How many different arrangements are possible? (ii) How many arrangements begin with T and end with a vowel?

Solution:

(i) There are 7 different letters in TUESDAY. $\text{Number of arrangements} = 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$

(ii) Using the box method for the restrictions:

First box: 1 choice (must be T)
Last box: 3 choices (vowels U, E, A)
Remaining 5 boxes: Can be filled in $5 \times 4 \times 3 \times 2 \times 1$ ways

$\text{Number of arrangements} = 1 \times 5 \times 4 \times 3 \times 2 \times 1 \times 3 = 360$

Worked example 3: Objects staying together

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Worked Example: Letters Staying Together

Question: In how many ways can the letters of the word NUMBERS be arranged if the letters M and B are always together?

Solution: When objects must stay together, we treat them as one unit.

(i) Total arrangements of NUMBERS = $7! = 5040$

(ii) If M and B must be together, treat MB as one unit:

We now have 6 units to arrange: N, U, E, R, S, (MB)
These 6 units can be arranged in $6!$ ways
Within the (MB) unit, M and B can be arranged in $2!$ ways

$\text{Number of arrangements} = 6! \times 2! = 720 \times 2 = 1440$

Common exam tips

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Strategic Approaches for Success:

Always identify restrictions first - deal with these before calculating remaining arrangements
Use the box method - it provides a clear visual approach
When objects must stay together, treat them as a single unit
Don't forget internal arrangements - if grouped objects can be rearranged within their group
Check your factorial calculations - these numbers grow very quickly

Remember!

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Key Points to Remember:

Permutations are arrangements where order matters
n distinct objects can be arranged in n! different ways
Use the box method - count choices for each position from left to right
Handle restrictions first - fill restricted positions before calculating remaining arrangements
Objects staying together should be treated as one unit, but don't forget internal arrangements within that unit

Arrangements (Permutations) (Leaving Cert Mathematics): Revision Notes