GCSE Maths revision · Higher: number and algebra

Counting and estimating roots

Higher only

The idea

The product rule for counting: if there are $a$ ways to do one thing and $b$ ways to do another, there are $a \times b$ ways to do both. To estimate a root, find the square (or cube) numbers either side.

Count the choices for each position, then multiply.
If items cannot repeat, the number of choices drops by one each time.
$\sqrt{50}$ lies between $\sqrt{49} = 7$ and $\sqrt{64} = 8$ , close to $7$ .

Watch out Multiply the choices, do not add them.

See it worked

Worked exampleCalculator allowed4 marks

A code is made from two letters of the alphabet (A to Z) followed by three digits (0 to 9).

(a) How many different codes are possible?

(b) How many codes are possible if the two letters must be different?

(a) Multiply the number of choices for each position. $26 \times 26 \times 10 \times 10 \times 10 = 676\,000$
(b) The second letter has only 25 choices.
$26 \times 25 \times 1000 = 650\,000$

Answer (a) $676\,000$ ; (b) $650\,000$

Your turn

Question 1Non-calculator3 marks

A café menu has 4 starters, 6 main courses and 3 desserts.

(a) How many different three-course meals (one starter, one main and one dessert) are possible?

(b) Lucy chooses a main course and exactly one other course (a starter or a dessert). How many different choices can she make?

Question 2Non-calculator4 marks

A three-digit number has no repeated digits and cannot start with 0.

How many such three-digit numbers are even?

Question 3Non-calculator4 marks

How many six-digit positive integers can be made using each of the digits 0, 1, 2, 3, 4 and 5 exactly once, if the integer must be divisible by 4? A six-digit integer cannot begin with 0.

In past papers

43 official questions in 41 papers since 2017. The latest:

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