In a combinatorics problem we first decide whether the choices are consecutive or alternative: for consecutive ones we multiply, for alternatives we add. Does order matter? If it does, we use permutations or arrangements without repetition (like 6 · 5 · 4); if not, Cₙᵏ. Two methods are often useful in problems with conditions. First: if two elements must stand next to each other, treat them as one block and permute inside the block. Second: for conditions such as “at least one”, subtract the number of unsuitable cases from the total. For example, choosing 3 people from 3 boys and 2 girls with at least one girl: C₅³ - C₃³ = 10 - 1 = 9.
Worked examples
5 pupils line up with two of them always next to each other: the block together with 3 others gives 4! = 24 ways, and 2 ways inside the block. Total 24 · 2 = 48.
Even three-digit numbers from the digits 1 to 6 (repetition allowed): the last digit is 2, 4 or 6, 3 ways, and the other two places 6 · 6. Total 6 · 6 · 3 = 108.
Class activity
“Lining up”: five pupils line up at the board. The class invents conditions (two pupils next to each other, a pupil at the end) and discusses how to count them, then tries out different orders.
Practice
1
In how many orders can 5 pupils line up if two of them always stand next to each other?
48
2
In how many ways can 3 people be chosen from 3 boys and 2 girls with at least one girl?
9
3
How many even three-digit numbers can be made from the digits 1 to 6 if repetition is allowed?
108
4
Why is it convenient to subtract the unsuitable cases from the total when the condition is “at least one”?
The unsuitable case is usually one simple case (no girl at all), and counting it is easier than counting the many variants of “at least one”.