Example 1: How many numbers can be made if all of the numbers $1, 2, 3, 4, 5$ are arranged in a line without repetition?

$5 \times 4 \times 3 \times 2 \times 1 = 120$

(Number of possibilities for each space.)

Thus:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Example 2: How many different "words" can be made using the letters $A, B, C, D, E, F, G, H$?

$8! = 40,320$

Example: How many numbers can be made by rearranging "$1234$" without using the same digit twice (other than $1$)?

The "long way": If the digit '$1$' is fixed in the first position, there are $3!$ ways of arranging the remaining digits.

$3! = 6 \quad (\text{ways of arranging the remaining 3 spaces})$

Since:

$3! \times 10 \quad (\text{ways of arranging the spaces}) = 60$

$(As\ opposed\ to\ 5! = 120\ for\ 12345 ).$

Example: How many "words" can be created from the following letters: "SHANNONABBOTT"?

Example: How many numbers can be created from "11234" that are above 32,000?

We start by considering the first digit:

Thus:

• For even slightly different questions, draw out possibilities. Therefore:

$12 + 3 + 3 = 18$

Example: How many odd numbers can be made from "12345"?

For the last digit to be odd (1, 3, 5):

$3 \times 4! = 72$

Example: Arranging numbers or letters, like multiplying a fixed number of matrices.

Example: Choosing a team from a squad or ordering a meal from a takeaway.

Example: From the numbers "$12345$", three different numbers are chosen. How many different numbers can be created in this case?

1. Ask: Is the order important? → Yes.
2. Draw out possibilities.
3. Calculate: $5 \times 4 \times 3 = 60 \text{ possibilities}$ $^5P_3 = 60 \quad \text{(Note:} ^nP_r = \frac{n!}{(n - r)!} \text{)}$

Example: How many "words" are possible when three letters from the alphabet are arranged randomly?

$26 \times 25 \times 24 \ {or}\ ^{26}P_3 \quad \text{or} \quad \frac{26!}{23!} = 15,600$

Example: A team of $11$ is chosen from a squad of $20$. Assuming all players can play all positions, how many possible teams are there?

Strategy: Classify players as "Chosen" or "Not Chosen".

• You will have $11C$ and $9NC$. C C C C C C C C C C C NC NC NC NC NC NC NC NC NC

Considering arrangement and repetitions:

$\frac{20!}{11! 9!} \quad \text{(Total arrangements, divide out repeated because order is not important.)}$

Thus:

$^{20}C_{11} = 167,960$