| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Specific items separated |
| Difficulty | Moderate -0.3 This is a standard permutations question with repeated letters and basic probability. Part (i) is routine factorial division (9!/4!2!), part (ii) uses complement counting with straightforward block arrangements, and part (iii) is systematic case-by-case counting. All techniques are textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(90720\) | B1 [1] | Not \(9!/2!2!\) |
| (ii) \(3\) vowels together \(= 3! \text{ or seen multiplied by integer or } 7 \text{ or } 6! \text{ seen multiplied as a num}\) | B1, B1 | |
| \(\text{Prob(not together)} = \frac{90720 - 7560}{90720} = \frac{83160}{90720} = 0.917 (= 11/12)\) | M1, A1 [4] | Subt from their (i) or dividing by their (i) or \(1 -\) prob; Correct answer from correct working |
| (iii) One S in \({}^6C_3\) ways \(= 10\) SS in \({}^5C_2\) ways \(= 10\) Total \(= 20\) | M1, M1, A1 [3] | \({}^6C_3\) seen added; \({}^5C_2\) seen added; Correct answer |
| OR \({}^6C_3 = {}^6C_3 \times 2\) or \({}^2\) or \(\times 1\) seen \({}^6C_3\) only | M1, M1, A1 | Correct answer |
**(i)** $90720$ | B1 [1] | Not $9!/2!2!$
**(ii)** $3$ vowels together $= 3! \text{ or seen multiplied by integer or } 7 \text{ or } 6! \text{ seen multiplied as a num}$ | B1, B1 |
$\text{Prob(not together)} = \frac{90720 - 7560}{90720} = \frac{83160}{90720} = 0.917 (= 11/12)$ | M1, A1 [4] | Subt from their (i) or dividing by their (i) or $1 -$ prob; Correct answer from correct working
**(iii)** One S in ${}^6C_3$ ways $= 10$ SS in ${}^5C_2$ ways $= 10$ Total $= 20$ | M1, M1, A1 [3] | ${}^6C_3$ seen added; ${}^5C_2$ seen added; Correct answer
OR ${}^6C_3 = {}^6C_3 \times 2$ or ${}^2$ or $\times 1$ seen ${}^6C_3$ only | M1, M1, A1 | Correct answer4 (i) Find the number of different ways that the 9 letters of the word HAPPINESS can be arranged in a line.\\
(ii) The 9 letters of the word HAPPINESS are arranged in random order in a line. Find the probability that the 3 vowels (A, E, I) are not all next to each other.\\
(iii) Find the number of different selections of 4 letters from the 9 letters of the word HAPPINESS which contain no Ps and either one or two Ss.
\hfill \mbox{\textit{CAIE S1 2011 Q4 [8]}}