Counting Principles, Arrangements and Selections

In short: Counting principles find how many outcomes there are without listing them all. If one choice has $m$ ways and the next $n$ ways, together they have $m \times n$ ways. Use $^{n} P_{r}$ when order matters and $^{n} C_{r}$ when it does not.

Starting point: Years 9 and 10

The multiplication principle

If one choice can be made in $m$ ways and a second in $n$ ways, together they can be made in $m \times n$ ways. This works for any number of choices.

Factorials

$n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1$

So $4! = 24$ and $5! = 120$ . By definition, $0! = 1$ .

$n!$ counts the ways to arrange $n$ different objects in a row. Five friends can stand in a line in $5! = 120$ ways.

Arrangements: order matters

Choosing $r$ objects from $n$ in order (like first, second, third):

$^{n} P_{r} = \frac{n !}{( n - r )!}$

Restrictions. Fill the restricted places first. How many 4-digit numbers can be made from 1, 2, 3, 4, 5 with no repeats, if the number must be even? The last digit is 2 or 4 (2 ways), then the rest: $4 \times 3 \times 2$ . Total $2 \times 24 = 48$ .

Selections: order does not matter

Choosing $r$ objects from $n$ when order does not matter (a group, a committee, a hand of cards):

$^{n} C_{r} = (r n) = \frac{n !}{r ! ( n - r )!}$

Counting

Multiplication principle

m ways then n ways: m × n

Factorials

n! arrangements in a row
0! = 1

Arrangements

Order matters
nPr = n!/(n - r)!
Restricted places first

Selections

Order does not matter
nCr = n!/(r!(n - r)!)

Probability

Favourable count over total count

Year 11Counting in probability

When all outcomes are equally likely,

$P = \frac{number of favourable outcomes}{total number of outcomes}$

and both counts can come from counting rules.

Example. Five people line up at random. What is the chance that Ana is first? Total $5! = 120$ . Ana first: $4! = 24$ . So $P = \frac{24}{120} = \frac{1}{5}$ .

Example. A bag has 5 red and 4 blue balls. Two are taken. $P (both red) = \frac{( 2 5 )}{( 2 9 )} = \frac{10}{36} = \frac{5}{18}$ , the same as the tree diagram answer $\frac{5}{9} \times \frac{4}{8}$ .

Identical objects. The letters of BOOK can be arranged in $\frac{4 !}{2 !} = 12$ ways, because swapping the two O's changes nothing.

Keeping two people together. Glue them into one block. Five people in a row with two friends together: arrange 4 items (the block and 3 others) in $4!$ ways, times $2!$ for the order inside the block. So $P (together) = \frac{4 ! \times 2 !}{5 !} = \frac{48}{120} = \frac{2}{5}$ .

Cards. A standard pack has 52 cards: 4 suits of 13, including 4 aces. The chance three cards dealt are all hearts is $\frac{( 3 13 )}{( 3 52 )} = \frac{286}{22 100} \approx 0.013$ .

At least one, with counting. The chance a 5-card hand has at least one ace is $1 - \frac{( 5 48 )}{( 5 52 )}$ , because the hands with no aces choose all 5 from the 48 other cards.

Handshakes and pairs. Each handshake is a pair of people with no order, so 10 people make $(2 10) = 45$ handshakes. A lottery choosing 6 numbers from 45 has $(6 45) = 8145060$ possible tickets.

Year 12Counting and random variables

Counting rules build probability distributions. Toss 4 coins and let $X$ be the number of heads. There are $2^{4} = 16$ outcomes, and $(k 4)$ of them have $k$ heads:

$x$	0	1	2	3	4
$P (X = x)$	$\frac{1}{16}$	$\frac{4}{16}$	$\frac{6}{16}$	$\frac{4}{16}$	$\frac{1}{16}$

The mean is $E (X) = 2$ , as you would expect. Mathematics Extension 1 takes this further with the binomial distribution.

Quick summary

Multiplication principle: $m$ ways then $n$ ways gives $m \times n$ .
$n! = n (n - 1) \dots 1$ arranges $n$ objects in a row. $0! = 1$ .
Order matters: $^{n} P_{r} = \frac{n !}{( n - r )!}$ . Fill restricted places first.
Order does not matter: $^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$ .
Probability = favourable count over total count.

Quiz

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A café offers 4 mains and 3 desserts. How many main and dessert meals are possible?
1. A 7
2. B 12
3. C 24
4. D 64
Show answer
B: 12
$4 \times 3$ .
What is $5!$ ?
1. A 15
2. B 25
3. C 120
4. D 60
Show answer
C: 120
$5 \times 4 \times 3 \times 2 \times 1$ .
What is $0!$ ?
1. A 0
2. B 1
3. C undefined
4. D 10
Show answer
B: 1
By definition $0! = 1$ .

Counting Principles, Arrangements and Selections

Starting point: Years 9 and 10

The multiplication principle

Factorials

Arrangements: order matters

Selections: order does not matter

Quick summary

Quiz

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