| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with grouped categories |
| Difficulty | Moderate -0.8 This is a straightforward permutations question testing standard techniques: treating grouped items as a single unit (part a(i)), fixing positions with restrictions (part a(ii)), and basic combinatorics with constraints (part b). All methods are textbook exercises requiring only direct application of formulas with no novel problem-solving or insight needed. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| (AAAIU) \(* * * *\), arrangements of vowels/repeats \(\times\) arrangements of (consonants & vowel group) | M1 | \(k \times 5!\) (\(k\) is an integer, \(k \geqslant 1\)) |
| \(\frac{5! \times 5!}{3!}\) | M1 | \(\frac{m}{3}!\) (\(m\) is an integer, \(m \geqslant 1\)). Both Ms can only be awarded if expression is fully correct |
| \(= 2400\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| E.g. R \(* * *\) T \(* * *\) L. Arrangements of consonants RL, RS, \(SL = {}^3P_2 = 6\). Arrangements of remaining letters \(= \frac{6!}{3!} = 120\) | M1 | \(k \times \frac{6!}{3!}\) or \(k \times {}^3P_2\) or \(k \times {}^3C_2\) or \(k \times 3!\) or \(k \times 3 \times 2\) (\(k\) is an integer, \(k \geqslant 1\)), no irrelevant addition |
| Total \(120 \times 6\) | M1 | Correct unsimplified expression or \(\frac{6!}{3!} \times {}^3C_2\) |
| \(= 720\) ways | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(1\quad 0\quad 3 = 2\times1\times4 = 8\) | M1 | Multiply 3 combinations, \({}^2C_x \times {}^8C_y \times {}^4C_z\). Accept \({}^2C_1 = 2\) etc. |
| 3 or more options correct unsimplified | A1 | |
| Summing their values of 4 or 5 legitimate scenarios | M1 | Summing *their* values of 4 or 5 legitimate scenarios (no extra scenarios) |
| Total \(= 366\) ways | A1 | Correct answer |
| Method 2: \({}^{14}C_4 - (2\text{N2R or 1N3R or 4R or 3R1B or 2R2B or 1R3B or 4B})\) | M1 | \({}^{14}C_4 - k\) seen, \(k\) an integer from an expression containing \({}^8C_x\) |
| \(1001-(1\times{}^8C_2+2\times{}^8C_3+{}^8C_4+{}^8C_3\times4+{}^8C_2{}^4C_2+8\times4+1)\) | A1 | 4 or more 'subtraction' options correct unsimplified |
| \(1001-(28+112+70+224+168+32+1)\) | M1 | *Their* \({}^{14}C_4\) – [*their* values of 6 or more legitimate scenarios] |
| \(= 366\) | A1 | Correct answer |
## Question 6(a)(i):
(AAAIU) $* * * *$, arrangements of vowels/repeats $\times$ arrangements of (consonants & vowel group) | M1 | $k \times 5!$ ($k$ is an integer, $k \geqslant 1$)
$\frac{5! \times 5!}{3!}$ | M1 | $\frac{m}{3}!$ ($m$ is an integer, $m \geqslant 1$). Both Ms can **only** be awarded if expression is **fully correct**
$= 2400$ | A1 | Correct answer
---
## Question 6(a)(ii):
E.g. R $* * *$ T $* * *$ L. Arrangements of consonants RL, RS, $SL = {}^3P_2 = 6$. Arrangements of remaining letters $= \frac{6!}{3!} = 120$ | M1 | $k \times \frac{6!}{3!}$ or $k \times {}^3P_2$ or $k \times {}^3C_2$ or $k \times 3!$ or $k \times 3 \times 2$ ($k$ is an integer, $k \geqslant 1$), no irrelevant addition
Total $120 \times 6$ | M1 | Correct unsimplified expression or $\frac{6!}{3!} \times {}^3C_2$
$= 720$ ways | A1 | Correct answer
---
## Question 6(b):
**Method 1:** N(2) R(8) Br(4):
$1\quad 2\quad 1 = 2\times{}^8C_2\times4 = 224$
$2\quad 1\quad 1 = 1\times{}^8C_1\times4 = 32$
$1\quad 1\quad 2 = 2\times8\times{}^4C_2 = 96$
$2\quad 0\quad 2 = 1\times1\times{}^4C_2 = 6$
$1\quad 0\quad 3 = 2\times1\times4 = 8$ | M1 | Multiply 3 combinations, ${}^2C_x \times {}^8C_y \times {}^4C_z$. Accept ${}^2C_1 = 2$ etc.
3 or more options correct unsimplified | A1 |
Summing their values of 4 or 5 legitimate scenarios | M1 | Summing *their* values of 4 or 5 legitimate scenarios (no extra scenarios)
Total $= 366$ ways | A1 | Correct answer
**Method 2:** ${}^{14}C_4 - (2\text{N2R or 1N3R or 4R or 3R1B or 2R2B or 1R3B or 4B})$ | M1 | ${}^{14}C_4 - k$ seen, $k$ an integer from an expression containing ${}^8C_x$
$1001-(1\times{}^8C_2+2\times{}^8C_3+{}^8C_4+{}^8C_3\times4+{}^8C_2{}^4C_2+8\times4+1)$ | A1 | 4 or more 'subtraction' options correct unsimplified
$1001-(28+112+70+224+168+32+1)$ | M1 | *Their* ${}^{14}C_4$ – [*their* values of 6 or more legitimate scenarios]
$= 366$ | A1 | Correct answer
---6
\begin{enumerate}[label=(\alph*)]
\item Find the number of ways in which all 9 letters of the word AUSTRALIA can be arranged in each of the following cases.
\begin{enumerate}[label=(\roman*)]
\item All the vowels (A, I, U are vowels) are together.
\item The letter T is in the central position and each end position is occupied by one of the other consonants (R, S, L).
\end{enumerate}\item Donna has 2 necklaces, 8 rings and 4 bracelets, all different. She chooses 4 pieces of jewellery. How many possible selections can she make if she chooses at least 1 necklace and at least 1 bracelet?
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2018 Q6 [10]}}