Permutations and combinations questions look different from every other topic: there is very little to calculate and a lot to decide. The hardest questions in Paper 5 (S1) and in Additional Math are almost always the ones where the counting has several conditions. This guide follows the order of the S1 revision workshop on our channel and shows how to organise those questions.
The one decision that comes first
Ask whether order matters. If you are arranging, ranking or forming a code or number, order matters, and you are counting arrangements (permutations). If you are choosing a team, a committee or a group, order does not matter, and you are counting selections (combinations). Words such as select, choose, make a group and form a team signal combinations. Say which it is in words before you write any numbers.
Strategy 1: Draw boxes and fill the restricted ones first
For arrangements with conditions, draw one box for each position and fill the boxes that have a condition first. For a number that must be even, fill the units digit first. If the number of choices for a later box depends on what went into an earlier box, as with “odd and greater than 500”, split into cases and add them. A useful rule: an or between cases means add, and an and between steps means multiply.
Strategy 2: Items together
Items that must be together are merged into a single box. Count the arrangements of the boxes, then multiply by the number of ways to arrange the items inside the merged box. With repeated letters you divide in both places, once for the outside boxes and once for the inside.
Strategy 3: Items apart
There are two methods, and it matters which one the wording points to.
- Total minus together. Count all arrangements and subtract those where the items are together. This is the right method for “not all of these are next to each other”.
- The gaps method. Arrange the other items first, then place the items to be separated into the gaps between them, including the two ends. This is the method for “none of these are next to each other”, because total minus together would need several awkward cases. With two items to separate, either method works.
The two ideas can be combined, for example one pair together and a group of others kept apart: merge the pair into a single item, then use the gaps. A case like “exactly two of three identical letters together” is solved by putting the pair in one box and the single letter in another, then separating those two boxes with the gaps method.
Strategy 4: Repeated items and alternating patterns
When some of the objects are identical, as with the letters of a word, divide by the factorial of each repeated count. For consonants and vowels in alternate positions, fix whether the arrangement starts with a consonant or a vowel (the question usually says), then arrange the consonants and the vowels separately and multiply. When the question fixes the first and last items to be the same type, make a case for each type and arrange only the middle.
Practice this topic
Ready to try some? Work through past paper questions on this topic one at a time. Attempt each question fully first, and open the mark scheme only afterwards.
Strategy 5: Permutation and combination notation
Arranging r of n objects is nPr, and selecting r of n is nCr. They are linked: nPr is the selection followed by an arrangement of the chosen items. If the number of objects equals the number of places, it is simply n factorial.
The link is the key to many exam questions. When an arrangement has a condition on who is selected, such as exactly three boys and two girls arranged in a line, or a particular card that must be included, a direct permutation will not work. Deal with the selection first using combinations, then arrange.
Strategy 6: Selections with repeated items
Combinations have no direct way to handle identical objects, so make cases for each repeated letter, for example none, one or two of a letter. You do not choose between identical copies, because it does not matter which one is selected. If the question says “at least” for the repeated letters, set those aside first. When the remaining letters are all different, the answer can be a single combination.
When selecting from groups with a rule such as “at least 11 of one kind, and equal numbers of the other two”, list the possible splits in a short table, work out the selections for each row, multiply across the groups, then add the rows.
Connection to probability
The probability is the number of favourable selections divided by the total number of selections. Count both with the same method, so the two are consistent. There is one important exception: when a probability question has combinations and repeated items together, treat all of the objects as different. Otherwise the outcomes are not equally likely and the favourable over total formula fails. In that case, unlike a pure counting question, you do choose among the repeated items.
Dividing into groups
Choose one group, then the next from those remaining. When the groups are of different sizes, that is the answer. When some groups have the same size, the groups are indistinguishable, so divide by the factorial of the number of equal-sized groups to remove double counting. For a probability that two named people end up in the same group, make a case for each group they could be in.
Habits that protect the marks
- Write the case structure before any numbers. A question worth 5 or 6 marks usually has two or three cases.
- Check a small version. If you are unsure, try the same idea with fewer objects and count by listing.
- Read the wording twice. “Exactly”, “at most”, “at least”, “not all” and “none” each change the count.
- Do not simplify too early. Leave products of combinations visible until the end, which makes your method easy to follow.
- Test that your answer is sensible. It should be positive and less than the total count.