Permutations and combinations for IGCSE Mathematics - Additional: Methods and Worked Examples

Arrangement versus selection: the core distinction

Permutations and combinations reduce to a single question: does the order of the chosen items matter? If rearranging the same items produces a different outcome, you need permutations. If the same group of items counts once regardless of the sequence in which you picked them, you need combinations. Every IGCSE Additional Mathematics problem on this topic rests on that distinction, and getting it wrong at the outset sends the entire solution down the wrong path.

This article covers three things. First, the factorial notation that underpins both formulas. Second, the permutation formula and its applications. Third, the combination formula and how to recognise when it applies. Each section includes worked examples drawn from the style of Cambridge 0606 papers.

Factorial notation

The symbol n! (read "n factorial") means the product of all positive integers from 1 to n:

• 5! = 5 x 4 x 3 x 2 x 1 = 120
• 3! = 3 x 2 x 1 = 6
• 1! = 1
• 0! = 1 (by definition)

The definition 0! = 1 is not arbitrary. It ensures the combination formula produces sensible results at its boundaries. For instance, choosing all n items from n items should give exactly 1 way, and nCn = n! / (n! x 0!) only equals 1 if 0! = 1.

Factorials grow rapidly. 10! = 3,628,800. This explosive growth is precisely why permutation and combination problems yield large answers even for modest values of n, and why the formulas exist: counting by hand becomes impractical almost immediately.

Permutations: when order matters

A permutation is an ordered arrangement. Choosing a president, vice-president, and secretary from a group of 10 people is a permutation problem because assigning Alice as president and Bob as secretary is different from assigning Bob as president and Alice as secretary.

The number of permutations of r items chosen from n distinct items is:

nPr = n! / (n - r)!

Notice the pattern in the calculation column. nPr always simplifies to the product of r consecutive descending integers starting from n. This shortcut avoids computing full factorials: 10P4 = 10 x 9 x 8 x 7, with no need to evaluate 10! or 6! separately.

Worked example 1: Arranging letters

How many different 4-letter arrangements can be formed from the letters A, B, C, D, E, F, G if no letter may be repeated?

1. There are 7 distinct letters and we want ordered arrangements of 4. This is a permutation: n = 7, r = 4.
2. 7P4 = 7! / (7 - 4)! = 7! / 3! = (7 x 6 x 5 x 4 x 3!) / 3! = 7 x 6 x 5 x 4 = 840

There are 840 possible arrangements.

Worked example 2: Filling positions

A committee of 12 members must elect a chairperson, a deputy chairperson, and a treasurer. No person may hold more than one position. In how many ways can these three positions be filled?

1. Three distinct roles drawn from 12 people. The roles are different, so order matters: this is a permutation with n = 12, r = 3.
2. 12P3 = 12 x 11 x 10 = 1320

Permutations of all n items

When you arrange all n items (r = n), the formula simplifies to nPn = n!. Arranging 5 books on a shelf: 5! = 120 ways. Arranging 8 athletes in a race: 8! = 40,320 orderings.

Combinations: when order does not matter

A combination is an unordered selection. Choosing 3 students from a class of 20 to form a study group is a combination problem: the group {Alice, Bob, Charlie} is the same group as {Charlie, Alice, Bob}.

The number of combinations of r items chosen from n distinct items is:

nCr = n! / (r! x (n - r)!)

The difference from the permutation formula is the extra r! in the denominator. That factor cancels out the duplicate arrangements of the same selection. For every group of r items, there are r! ways to arrange them internally, and dividing by r! collapses all of those into a single selection.

A useful symmetry: nCr = nC(n-r). Choosing 3 items from 10 produces the same count as choosing 7 items from 10, because every selection of 3 items to include is simultaneously a selection of 7 items to exclude. This identity can simplify calculations: 10C8 = 10C2 = 45, which is far quicker to compute than evaluating 8! directly.

Worked example 3: Selecting a team

A squad of 15 football players must have 11 selected for a match. In how many ways can the 11 players be chosen?

1. Selection with no role assignment. Order does not matter: this is a combination with n = 15, r = 11.
2. Using the symmetry property: 15C11 = 15C4 = (15 x 14 x 13 x 12) / (4 x 3 x 2 x 1) = 32,760 / 24 = 1365

Worked example 4: Mixed selection with conditions

A panel of 5 must be chosen from 6 men and 4 women. The panel must include at least 2 women. How many different panels are possible?

1. "At least 2 women" means 2, 3, or 4 women on the panel. Calculate each case separately.
2. Case 1: 2 women and 3 men. 4C2 x 6C3 = 6 x 20 = 120
3. Case 2: 3 women and 2 men. 4C3 x 6C2 = 4 x 15 = 60
4. Case 3: 4 women and 1 man. 4C4 x 6C1 = 1 x 6 = 6
5. Total = 120 + 60 + 6 = 186

The complementary approach also works: total panels without restriction (10C5 = 252) minus panels with 0 women (6C5 = 6) minus panels with exactly 1 woman (4C1 x 6C4 = 60), giving 252 - 6 - 60 = 186. Both routes reach the same answer, and the complementary method is sometimes faster when "at least" involves many valid cases.

Distinguishing permutation from combination problems

The Cambridge IGCSE Additional Mathematics papers rarely label a question as "use permutations" or "use combinations." You must decide from context. The following decision framework works reliably.

Problems involving both permutations and combinations

Some IGCSE exam questions require both formulas in the same solution. A typical structure: first select a group (combination), then arrange the selected items (permutation).

Worked example 5: Select then arrange

From 8 different books, 4 are to be chosen and placed on a shelf in a row. Find the number of possible arrangements.

1. Method 1 (direct): This is simply 8P4 = 8 x 7 x 6 x 5 = 1680, because we are choosing and ordering in one step.
2. Method 2 (two-stage): First choose 4 from 8 (8C4 = 70), then arrange the 4 chosen books (4! = 24). Total = 70 x 24 = 1680.

Both methods give 1680. Method 2 is instructive because it separates the selection stage from the arrangement stage, making the logic transparent. On exam papers, Method 2 is often the safer approach for complex problems because each stage can be checked independently.

Worked example 6: Conditional arrangement

Five people are to sit in a row of 7 chairs. In how many ways can this be done?

1. First choose which 5 of the 7 chairs will be occupied: 7C5 = 21 ways.
2. Then assign the 5 people to those 5 chairs: 5! = 120 ways.
3. Total = 21 x 120 = 2520.

Alternatively, think of it as choosing a chair for each person in sequence. Person 1 has 7 choices, person 2 has 6, down to person 5 with 3 choices: 7 x 6 x 5 x 4 x 3 = 2520. Same result, different reasoning.

Common errors and how to avoid them

Self-check questions

1. Evaluate 7P3 and 7C3. Verify that 7P3 = 7C3 x 3!.
2. A lock has a 4-digit code using digits 0-9 with no repetition. How many codes are possible?
3. A group of 9 students must split into one group of 4 and one group of 5. How many ways can this be done?
4. From 7 boys and 5 girls, a team of 6 is chosen. The team must include at least 3 girls. Find the number of possible teams.
5. Eight athletes compete in a race. In how many ways can the gold, silver, and bronze medals be awarded?

Exam strategy

Cambridge examiners consistently reward clear structure. Write down whether the problem is a permutation or a combination before starting the calculation. State the values of n and r explicitly. Show the formula with numbers substituted, then evaluate. This habit takes seconds and eliminates the most common source of lost marks: using the wrong formula without realising it.

For problems that combine selection with arrangement, separate the stages clearly on paper. Label them "Selection" and "Arrangement" or "Step 1" and "Step 2". Multiply the results of independent stages. The examiners on IGCSE Additional Mathematics papers award method marks for correct intermediate steps even if the final answer contains an arithmetic slip, so showing the structure of your reasoning has genuine mark value.

When a question says "at least" or "at most," consider whether the direct approach (summing valid cases) or the complementary approach (total minus invalid cases) involves fewer calculations. If "at least 1" appears, the complement is almost always faster: total ways minus the ways with none.