Arrangements (VCE SSCE Mathematical Methods): Revision Notes

Arrangements

What are arrangements?

An arrangement is a way of ordering a set of objects where the sequence matters. Each different order counts as a distinct arrangement, even when using the same objects.

For example, if we arrange the three letters A, B, C, we can create six different arrangements:

ABC, ACB, BAC, BCA, CAB, CBA

Notice that ABC and CBA are considered different arrangements because the order of the letters is different. In mathematics, arrangements are also called permutations.

The box method for counting arrangements

When we have more than three objects, listing all possible arrangements becomes impractical. Instead, we can use a visual box method to work out how many arrangements exist.

Arranging three letters

Let's revisit arranging the letters A, B, C using the box method. We imagine three empty boxes that need to be filled:

Now we count the choices available for each box:

• We have 3 choices for the first box (we can pick A, B, or C)
• We have 2 choices for the second box (we've already used one letter)
• We have 1 choice for the third box (we've already used two letters)

Using the multiplication rule, the total number of arrangements is:

$$3 \times 2 \times 1 = 6$$

This matches the six arrangements we listed earlier!

Worked example: arranging four books

Worked example: arranging twelve students

Factorial notation

Writing out long multiplication sequences like $12 \times 11 \times 10 \times \ldots \times 2 \times 1$ is cumbersome, so mathematicians use a shorthand called factorial notation.

The symbol $n!$ (read as "n factorial") represents the product of all positive integers from $n$ down to 1:

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

For example:

• $3! = 3 \times 2 \times 1 = 6$
• $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
• $12! = 12 \times 11 \times 10 \times \cdots \times 2 \times 1 = 479,001,600$

Arranging some objects from a larger set

Sometimes we want to arrange only some objects from a larger collection. For instance, how many ways can we arrange two letters chosen from A, B, C?

The possibilities are: AB, BA, AC, CA, BC, CB (six arrangements)

Using the box method with two boxes:

We have:

• 3 choices for the first box (A, B, or C)
• 2 choices for the second box (one letter already used)

Total arrangements: $3 \times 2 = 6$

Worked example: painting the Olympic rings

The general formula

We can express the answer from the Olympic rings example using factorials. We calculated:

$$8 \times 7 \times 6 \times 5 \times 4$$

If we multiply by 1 (which doesn't change the value), we can write:

$$8 \times 7 \times 6 \times 5 \times 4 = (8 \times 7 \times 6 \times 5 \times 4) \times \frac{3 \times 2 \times 1}{3 \times 2 \times 1} = \frac{8!}{3!}$$

Notice that $3 = 8 - 5$, so this can also be written as:

$$\frac{8!}{(8-5)!}$$

Worked example: forming four-digit numbers

The value of 0!

For our arrangement formula to work in all situations, we need to define $0!$.

Consider arranging $n$ objects in groups of size $n$ (using all objects). From first principles, this equals $n!$.

Using our formula:

$$^nP_n = \frac{n!}{(n-n)!} = \frac{n!}{0!}$$

For this to equal $n!$, we must have:

$$0! = 1$$

Arrangements with restrictions

When an arrangement has restrictions or conditions, the best strategy is to deal with the restriction first, then count the remaining choices.

Worked example: forming even four-digit numbers

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Remember!