SUBJECT: GRADE 12 MATHEMATICS PAPER 1 ERRATA
The word “repetition” and the fact that the word contains 2 x “R’s” contributed to the ambiguity of this question.
Below are solutions that may be considered for this examination ONLY.
QUESTION 11.2
11.2 Consider the letters of the word: NUMERATOR.
11.2.1 How many 9 letter word-arrangements can be formed, if repetition of letters is allowed? (1)
11.2.2 How many 9 letter word-arrangements can be formed, if all 4 vowels are never together and repetition of letters is not allowed? (3)
11.2.3 An 8 letter word-arrangement is made from the word NUMERATOR. All the vowels must be included in this word-arrangement and repetition of
letters is not allowed. What is the probability that all odd-number spaces are occupied by vowels? (4) [15]
| ORIGINAL MARKING GUIDELINE | ||
| 11.2.1 | 99 or 387 420 489 | √ 99 (1) |
| 11.2.2 | If vowels are together: 6! × 4! ∴ if vowels are not together: 9! - (6! × 4!) = 345 600 | √ 6! × 4! √ subtracting from 9! √ answer (3) |
| 11.2.3 | Vowels in odd spaces: = 4 × 5 × 3 × 4 × 2 × 3 × 1 × 2 = (4 × 3 × 2 × 1) × (5 × 4 × 3 × 2) = 4! × 120 = 2880 ∴ probability = 2880 (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2) = 2880 362880 = 1 126 | √ 4! × 120 √ vowels in odd spaces (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2) √ answer (4) [15] |
| CORRECTION TO MARKING GUIDELINE | ||
| 11.2.1 | Since there are only 8 distinct letters, the answer should be: 89 or 134 217 728 | √ 89 (1) |
| 11.2.2 | if vowels are together: OR The ambiguity might lead to the understanding that there are only 8 distinct letters but the word must contain 9 letters. NO SOLUTION | √ 6! × 4! 2! √ 9! 2! √ answer (3) |
| 11.2.3 | Since there are only 4 distinct consonants: ∴ probability = 576 | √ 4! × 4! √ vowels in odd spaces (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) √ answer (4) [15] |
After considering the impact of the ambiguity of the question, schools are informed not to mark this sub-question 11.2 and mark out of 142.