The Fundamental Counting Principle

Q: What does 0! equal?

$0! = 1$. It makes formulae for arrangements work when nothing is left to arrange.

The fundamental counting principle lets you count huge numbers of outcomes without listing them. It is used for codes, number plates and arrangements, and then to calculate probabilities by dividing favourable outcomes by total outcomes.

Counting outcomes

If one choice can be made in $m$ ways and a second independent choice in $n$ ways, the two can be made together in $m \times n$ ways.

Draw a box for each position and write the number of options in each:

Example. A number plate has 3 letters followed by 3 digits, with repetition allowed:

$26 \times 26 \times 26 \times 10 \times 10 \times 10 = 17576000$

Without repetition, the options drop by one at each position. A 4-digit code using digits 0 to 9 with no repeats:

$10 \times 9 \times 8 \times 7 = 5040$

Fill restricted positions first, then the rest.

Factorial notation

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

$n!$ counts the ways to arrange $n$ different objects in a row. By definition $0! = 1$ .

Example. 7 different books on a shelf: $7! = 5040$ arrangements.

If 2 particular books must be at the ends: they can swap ( $2$ ways), and the other 5 fill the middle ( $5!$ ways):

$2 \times 5! = 240$

Using counting to find probabilities

If all outcomes are equally likely:

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

Example. A 4-digit code is made from the digits 0 to 9 without repetition. What is the probability that the code starts with 9 and is even?

Total codes: $10 \times 9 \times 8 \times 7 = 5040$
Favourable: first digit 9 (1 way), last digit even (5 ways), middle digits $8 \times 7$

$P = \frac{1 \times 8 \times 7 \times 5}{5 040} = \frac{280}{5 040} = \frac{1}{18}$

Questions that need permutations or combinations ( $^{n} P_{r}$ , $^{n} C_{r}$ ) are not examinable. Use the fundamental counting principle and factorials.

Key terms

Fundamental counting principle: If choices can be made in $m$ and $n$ ways, together they can be made in $mn$ ways.
Factorial: $n! = n (n - 1) (n - 2) \dots 1$ .
Repetition: Whether an item or digit may be used more than once.
Equally likely: Every outcome has the same chance of occurring.

Common mistakes

Filling unrestricted positions first, leaving too few options for the restricted ones.
Allowing repetition when the question says without repetition, or the reverse.
Forgetting that 0 cannot start a number but can start a code, depending on the question.
Splitting into cases incorrectly when a restriction depends on another position.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

A number plate consists of 3 letters of the alphabet followed by 3 digits. Letters and digits may be repeated. How many different number plates are possible?

Worked solution

Hidden - try the question first

Solution

Letters: $26 \times 26 \times 26$
Digits: $10 \times 10 \times 10$
$2 6^{3} \times 1 0^{3} = 17576000$

Question 2

5 marks total

routine

Seven different books are arranged on a shelf.

(a) In how many ways can they be arranged? (2)

(b) In how many ways can they be arranged if two particular books must be at the two ends? (3)

Worked solution

Hidden - try the question first

Solution

(a) $26 \times 26 \times 26$
$= 5040$
(b) The two books at the ends: $2$ ways
The other 5 books: $5! = 120$ ways
$2 \times 120 = 240$

Question 3

6 marks total

complex

A 4-digit code is formed from the digits 0 to 9. Digits may not be repeated.

(a) How many codes are possible? (2)

(b) Calculate the probability that a code chosen at random starts with 9 and is even. (4)

Worked solution

Hidden - try the question first

Solution

(a) $26 \times 26 \times 26$
$= 5040$
(b) First digit: 1 way (9); last digit: 5 ways (0, 2, 4, 6, 8)
Middle digits: $8 \times 7 = 56$
Favourable $= 1 \times 56 \times 5 = 280$
$P = \frac{280}{5 040} = \frac{1}{18}$

Question 4

9 marks total

problem-solving

A cellphone PIN has 5 digits chosen from 1 to 9 (0 is not used). Digits may be repeated unless stated otherwise.

(a) How many PINs are possible? (1)

(b) How many PINs have no repeated digits? (2)

(d) How many PINs with no repeated digits are odd and greater than 50 000? (3)

Worked solution

Hidden - try the question first

Solution

(a) $26 \times 26 \times 26$
(b) $9 \times 8 \times 7 \times 6 \times 5$
$= 15120$
(c) $P (no repeats) = \frac{15 120}{59 049} \approx 0, 2561$
$P (at least one repeat) = 1 - 0, 2561$
$\approx 0, 7439$
(d) First digit 5, 7 or 9 (odd): last digit 4 ways, middle $7 \times 6 \times 5$ , so $3 \times 4 \times 210 = 2520$
First digit 6 or 8 (even): last digit 5 ways, so $2 \times 5 \times 210 = 2100$
Total $= 4620$

FAQs

Are combinations (nCr) in the IEB Grade 12 syllabus?

No. Questions that need permutations or combinations (nPr, nCr) are not examinable in the IEB exam. Use the counting principle and factorials.

What does 0! equal?

$0! = 1$ . It makes formulae for arrangements work when nothing is left to arrange.

How do I handle a restriction that depends on another position?

Split into cases. For example, an odd first digit uses up one of the odd options for the last digit, so count that case separately.

Counting outcomes

Factorial notation

Using counting to find probabilities

Key terms

Common mistakes

Practice questions

FAQs

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