The complement method counts the outcomes you do not want and subtracts them from the total: wanted = total − unwanted. It works well when “unwanted” is one simple case but “wanted” is a long list. Both selections and arrangements in permutations and combinations can use it.
This lesson builds on counting selections with a restriction, where similar answers were found by adding cases.
When should you reach for a complement?
Look for these phrases:
- at least one (the complement is none),
- not adjacent or not next to each other (the complement is adjacent),
- not all the same (the complement is all the same),
- at most with a large upper limit.
Follow three steps: count the total with no condition, count the unwanted outcomes, then subtract.
Worked example 1: a selection
A committee of 4 is chosen from 6 men and 5 women. Find the number of committees with at least one woman.
Total: 11C4 = (11 × 10 × 9 × 8) / 24 = 7920 / 24 = 330.
Unwanted (no women, so 4 men): 6C4 = 15.
Answer: 330 − 15 = 315.
Check (direct): 1 woman: 5 × 6C3 = 5 × 20 = 100. 2 women: 5C2 × 6C2 = 10 × 15 = 150. 3 women: 5C3 × 6 = 10 × 6 = 60. 4 women: 5C4 = 5. The sum is 100 + 150 + 60 + 5 = 315. The answers agree, and the complement took two lines instead of four cases.
Worked example 2: an arrangement
Six people stand in a row. Find the number of arrangements in which A and B are not next to each other.
Total: 6! = 720.
Unwanted (A and B adjacent): treat A and B as one block. There are then 5 objects, arranged in 5! = 120 ways, and A and B can swap inside the block in 2 ways. So 120 × 2 = 240.
Answer: 720 − 240 = 480.
Check (gap method): arrange the other 4 people in 4! = 24 ways. That leaves 5 gaps, and A and B go into two different gaps in 5 × 4 = 20 ways. So 24 × 20 = 480. The answers agree.
The mistake to watch for
A common slip is to use the wrong complement for “at least one”.
Question: Count the committees in example 1 with at least one woman.
Mistaken answer: 330 − 5C4 = 330 − 5 = 325
The student subtracted the committees with all women, not the committees with no women.
“At least one woman” fails only when there are no women, which means 4 men. The correction is 330 − 6C4 = 330 − 15 = 315.
Check yourself
1. A group of 5 is chosen from 4 boys and 5 girls. How many groups have at least one boy?
Show answer
Total 9C5 = 126. No boys means all 5 girls: 5C5 = 1. So 126 − 1 = 125.
2. Five people stand in a row. In how many ways are X and Y not next to each other?
Show answer
Total 5! = 120. Adjacent: 4! × 2 = 48. So 120 − 48 = 72.
Check by gaps: the other 3 people in 3! = 6 ways, 4 gaps, X and Y into different gaps in 4 × 3 = 12 ways, and 6 × 12 = 72.
3. A 4-digit code uses digits 0 to 9, and digits may be repeated. How many codes contain at least one 7?
Show answer
Total 10⁴ = 10 000. Codes with no 7 use 9 digits in each place: 9⁴ = 6561. So 10 000 − 6561 = 3439.
Where this leads next
The last skill in this module is checking whether a proposed method counts too much: explain overcounting in a proposed method. You can also try the mixed practice set. The non-calculator working trainer is useful for checking your factorial cancelling.
Students can understand the complement idea yet not notice when it applies. A teacher in online one-to-one Additional Mathematics tuition can help you build the habit of scanning the wording for those signal phrases.