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# Factorial Notation

## Worked example 12: The arrangement of outcomes without repetition

Eight athletes take part in a $$\text{400}$$ $$\text{m}$$ race. In how many different ways can all $$\text{8}$$ places in the race be arranged?

Any of the $$\text{8}$$ athletes can come first in the race. Now there are only $$\text{7}$$ athletes left to be second, because an athlete cannot be both second and first in the race. After second place, there are only $$\text{6}$$ athletes left for the third place, $$\text{5}$$ athletes for the fourth place, $$\text{4}$$ athletes for the fifth place, $$\text{3}$$ athletes for the sixth place, $$\text{2}$$ athletes for the seventh place and $$\text{1}$$ athlete for the eighth place. Therefore the number of ways that the athletes can be ordered is as follows: $8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = \text{40 320}$

As in the example above, it is a common occurrence in counting problems that the outcome of the first event reduces the number of possible outcomes for the second event by exactly $$\text{1}$$, and the outcome of the second event reduces the possible outcomes for the third event by $$\text{1}$$ more, etc.

As this sort of problem occurs so frequently, we have a special notation to represent the answer. For an integer, $$n$$, the notation $$n!$$ (read $$n$$ factorial) represents:

$$n\times \left(n-1\right)\times \left(n-2\right)\times \cdots \times 3\times 2\times 1$$

This allows us to formulate the following:

The total number of possible arrangements of $$n$$ different objects is $n \times (n-1) \times (n-2) \times \ldots \times 3 \times 2 \times 1 = n!$

with the following definition: 0! = 1.

## Worked example 13: Factorial notation

1. Determine $$12!$$
2. Show that $$\dfrac{8!}{4!} = 8 \times 7 \times 6 \times 5$$
3. Show that $$\dfrac{n!}{(n-1)!} = n$$

1. We know from the definition of a factorial that $$12! = 12 \times 11 \times 10 \times \ldots \times 3 \times 2 \times 1$$. However, it can be quite tedious to work this out by calculating each multiplication step on paper or typing each step into your calculator. Fortunately, there is a button on your calculator which makes this much easier. To use your calculator to work out the factorial of a number:

• Input the number.

• Press SHIFT on your CASIO or 2ndF on your SHARP calculator.

• Then press $$x!$$ on your CASIO or $$n!$$ on your SHARP calculator.

• Finally, press equals to calculate the answer.

If we follow these steps for $$12!$$, we get the answer $$\text{479 001 600}$$.

2. Expand the factorial notation: $\dfrac{8!}{4!} = \dfrac{8 \times 7 \times 6 \times 5 \times \cancel{4} \times \cancel{3} \times \cancel{2} \times \cancel{1}}{\cancel{4} \times \cancel{3} \times \cancel{2} \times \cancel{1}} = 8 \times 7 \times 6 \times 5 = \text{ RHS}$
3. Expand the factorial notation:$\dfrac{n!}{(n-1)!} = \dfrac{n \times \cancel{(n-1)} \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}}{\cancel{(n-1)} \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}} = n$

If $$n = 1$$, we get $$\frac{1!}{0!}$$. This is a special case. Both $$1!$$ and $$0! =1$$, therefore $$\frac{1!}{0!} = 1$$ so our identity still holds.

## Factorial notation

Exercise 10.5

Work out the following without using a calculator:

$$3!$$

$3 \times 2 \times 1 = \text{6}$

$$6!$$

$6 \times 5 \times 4 \times 3 \times 2 \times 1 = \text{720}$

$$2!3!$$

$2 \times 1 \times 3 \times 2 \times 1 = \text{12}$

$$8!$$

$8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = \text{40 320}$

$$\dfrac{6!}{3!}$$

$\dfrac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = \text{120}$

$$6! + 4! - 3!$$

$(6 \times 5 \times 4 \times 3 \times 2 \times 1) + (4 \times 3 \times 2 \times 1) - (3 \times 2 \times 1)= 720 + 24 -6 = \text{738}$

$$\dfrac{6! - 2!}{2!}$$

$\dfrac{(6 \times 5 \times 4 \times 3 \times 2 \times 1) - (2 \times 1)}{2 \times 1}= \dfrac{720 - 2}{2} = \text{359}$

$$\dfrac{2! + 3!}{5!}$$

$\dfrac{(2 \times 1) + (3 \times 2 \times 1)}{5 \times 4 \times 3 \times 2 \times 1}= \dfrac{2 + 6}{120} = \dfrac{1}{15}$

$$\dfrac{2! + 3! - 5!}{3! - 2!}$$

$\dfrac{2 + 6 - 120}{6 - 2}= \dfrac{-\text{112}}{4} = -\text{28}$

$$(3!)^{3}$$

$$6 \times 6 \times 6 = \text{216}$$

$$\dfrac{3! \times 4!}{2!}$$

$\dfrac{(3 \times 2 \times 1) \times (4 \times 3 \times 2 \times 1)}{2 \times 1} = \text{72}$

Calculate the following using a calculator:

$$\dfrac{12!}{2!}$$

$\dfrac{(12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}{2 \times 1} = \text{239 500 800}$

$$\dfrac{10!}{20!}$$

$$\text{1,49} \times \text{10}^{-\text{12}}$$

$$\dfrac{10! + 12!}{5! + 6!}$$

$$\text{574 560}$$

$$5!(2! + 3!)$$

$$\text{960}$$

$$(4!)^{2}(3!)^{2}$$

$$\text{20 736}$$

Show that the following is true:

$$\dfrac{n!}{(n-2)!} = n^{2} - n$$

\begin{align*} \dfrac{n!}{(n-2)!} & = \dfrac{n \times (n-1) \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}}{\cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}} \\ \\ &= n(n-1) = n^{2} - n \end{align*}

$$\dfrac{(n-1)!}{n!} = \dfrac{1}{n}$$

$\dfrac{(n-1)!}{n!} = \dfrac{\cancel{(n-1)} \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}}{n \times \cancel{(n-1)} \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}} = \dfrac{1}{n}$

$$\dfrac{(n-2)!}{(n-1)!} = \dfrac{1}{n-1} \text{ for } n>1$$

$\dfrac{(n-2)!}{(n-1)!} = \dfrac{\cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}}{(n-1) \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}} = \dfrac{1}{n-1}$