Grade 12 Maths · Probability · Count using the "not" rule

Count using the "not" rule Grade 12

Count arrangements that break a condition by taking the total and subtracting the ones that meet it, for example "not together" = all arrangements - together, or "at least one" = total - none.

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Grade 12 Permutations with restrictions arranging people that MUST NOT sit together (NSC Maths)|NTENTE \ New Technological Education

People Arrangements & The "Not Next To" Rule | Counting Principle Part 8 (Gr 12)The daily Dose of Mathematics

Find the number of 4-digit numbers containing at least one digit 5. The Complement PrincipleMs Shaws Math Class

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Practice · NavyBlue · 2 marks

Consider the letters of the word LETTER.

In how many of these arrangements are the two Ts NOT next to each other?(2)

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More questions: count using the "not" rule

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3. Question 11.2.2 · September 2026 · Paper 1 · Provincial · 3 marks
9 members of a school's Representative Council of Learners (RCL) are seated in the front row for the school's photographs.
11.2 The following office bearers/portfolio managers are present at the school's photo session:

• President (P)
• Deputy President (DP)
• Secretary (S) and
• Public Relations Officer (PRO)

Answer the questions that follow.
11.2.2 If the Secretary and Public Relations Officer may NOT sit together, in how many different ways can the RCL members be seated?(3)
Show the memo (3 ticks)Memo
1. ✓
  MethodTotal $-$ together
  1 mark · Count using the "not" rule
2. ✓
  $2! \times 8!$ $= 9! - (2! \times 8!)$
  1 mark · Count using the "not" rule
3. ✓
  Answer $= 282240$
  1 mark · Count using the "not" rule
4. Question 10.3.2 · May/June 2023 · Paper 1 · National (DBE) · 3 marks
10.3 Seven friends (4 boys and 3 girls) want to stand in a straight line next to each other to take a photo.

Answer the questions that follow.
10.3.2 In the next photo, determine the probability that Selwyn (a boy) and Lindiwe (a girl) will NOT stand next to each other in the photo.(3)
Show the memo (3 ticks)Memo
1. ✓
  METHOD 1: $7! - 6! \times 2$ $7! - 6! \times 2$
  1 mark · Count using the "not" rule
2. ✓
  METHOD 1: Denominator ( $7!$ ) $\frac{7 ! - 6 ! \times 2}{7 !}$
  1 mark · Use the counting principle
3. ✓
  METHOD 1: Answer $= \frac{5}{7} = 0, 71$
  1 mark · Calculate a simple probability
5. Question 11.3.2 · November 2025 · Paper 1 · National (DBE) · 4 marks
11.3 Eight runners compete in a race where there are no tied finishes. Bongi and Andrew are two of the competitors.

Answer the questions that follow.
11.3.2 Calculate the probability that TWO OR MORE runners finish the race after Andrew and before Bongi.(4)
Show the memo (4 ticks)Memo
1. ✓
  METHOD 1: $6! (15)$ Possible outcomes (A = Andrew, B = Bongi):
  $A \times 6 \times 5 \times B \times 4 \times 3 \times 2 \times 1$
  $A \times 6 \times 5 \times 4 \times B \times 3 \times 2 \times 1$
  $A \times 6 \times 5 \times 4 \times 3 \times B \times 2 \times 1$
  $A \times 6 \times 5 \times 4 \times 3 \times 2 \times B \times 1$
  $A \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times B$
  $5 \times 6! + 4 \times 6! + 3 \times 6! + 2 \times 6! + 1 \times 6!$
  $= 6! (5 + 4 + 3 + 2 + 1)$
  $= 6! (15)$
  2 marks · Count arrangements with a condition
2. ✓
  METHOD 1: $8!$ in denominator $P = \frac{\dots}{8 !}$
  1 mark · Use the counting principle
3. ✓
  METHOD 1: Answer $P (two or more runners after Andrew and before Bongi) = \frac{6 ! ( 15 )}{8 !} = \frac{15}{56} = 0, 27$
  1 mark · Calculate a simple probability
6. Practice · NavyBlue · 2 marks
Nine beads are threaded in a row: $4$ identical red beads, $3$ identical blue beads and $2$ identical white beads.
In how many of these orders are the two white beads NOT next to each other?(2)
Show the memo (2 ticks)Memo
1. ✓
  White beads togetherWW as one block: $\frac{8 !}{4 ! \times 3 !} = 280$
  1 mark · Count arrangements with repeated items
2. ✓
  Answer $1260 - 280 = 980$
  1 mark · Count using the "not" rule

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