IGCSE Additional Mathematics (0606)

Arrangement and selection problems

What this topic tests, the mistakes that cost marks, and one exam-style question worked through step by step.

Key ideas and common mistakes

Counting the number of ways to arrange a selection of items without repetition (permutations), recognizing that order matters and that the number of choices decreases with each selection.
Common mistake: A common error is to use combinations or allow repeats.
When groups must stay together in arrangements, treat each group as a single block to arrange the blocks, then multiply by the number of ways to arrange members within each block.
Common mistake: Students often forget to include the internal arrangements.
To count numbers above a threshold, restrict the first digit to those that make the number large enough, then permute the remaining digits.
Common mistake: The common mistake is to ignore the leading digit's role or allow repeats.
Counting selections with 'at least one' of a type is best done by subtracting the case with none of that type from the total, since directly adding cases (1, 2, 3, etc.) is slower and error-prone.
Fill the restricted first and last places first, remembering the first letter is already used so the last place has fewer choices, then multiply by the arrangements of the remaining letters.
When a restriction (like 'exactly two even digits') automatically ensures another condition ('at least two odd digits'), avoid overcounting by recognizing only one valid case, not splitting into unnecessary subcases.

Worked example

Seven different non-zero digits are available and no digit may be repeated. Find how many different 5-digit numbers can be formed when there are no restrictions on which of the digits are used.

Try it first, then show the working

Order matters here, because rearranging the digits produces a different number, and repetition is not allowed, so this is a permutation.
We are choosing and arranging 5 of the 7 available digits, which is 7P5.
7P5 = 7!/(7 - 5)! = 7 * 6 * 5 * 4 * 3 = 2520.

Answer: 2520

Try one yourself

The 11 musicians are placed in a single row for a photograph. Find the number of different rows that are possible if the violinists are all next to…Answer it and get it marked → Four of the digits 2, 3, 4, 5, 6, 7, 9 are arranged, with no digit repeated, to form a 4-digit number. Find how many of these numbers are greater…Answer it and get it marked → Priya takes 4 books from this shelf. Find the number of ways she can do this if at least 1 of the books taken is a novel.Answer it and get it marked →

21 exam-style questions on Arrangement and selection problems, each marked to the scheme.

Arrangement and selection problems

Key ideas and common mistakes

Worked example

Try one yourself

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