| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Code/password formation |
| Difficulty | Moderate -0.5 This is a straightforward permutations question with standard counting principles. Part (a) is direct application of the multiplication principle, part (b) uses inclusion-exclusion (a common technique), and part (c) combines counting with basic probability. While multi-part, each step follows textbook methods without requiring novel insight or complex reasoning. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(^5P_2 \times\ ^7P_4\) or \(5\times4\times7\times6\times5\times4\) | M1 | \(^5P_x \times\ ^7P_y,\ 1\leq x\leq 4,\ 1\leq y\leq 6\) |
| \(16800\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| With A and no 5: \(8\times\ ^6P_4\) or \((1\times4\times6\times5\times4\times3)\times2\) or \(4C1\times2!\times6P4 = 2880\) | M1 | One number of ways correct, accept unsimplified. |
| With 5 and no A: \(^4P_2\times4\times\ ^6P_3 = 5760\); With A and 5: \(8\times4\times\ ^6P_3 = 3840\) | M1 | Add 2 or 3 identified correct scenarios only, accept unsimplified. |
| Total \(= 12480\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| No A or 5: \((4\times3)\times(6\times5\times4\times3) = 4320\) | M1 | \(^4P_2 \times\ ^6P_4\) or \(^4C_2\times\ ^6C_4\) seen, accept unsimplified. |
| Required number \(= \textit{their}\ \textbf{(a)} - \textit{their}\ 4320\) | M1 | *Their* 5(a) (or correct) \(-\) *their* (No A or 5) value. |
| \(12480\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| With A: \(^4P_1\times\ ^7P_4\times2 = 6720\); With 5: \(^5P_2\times\ ^6P_3\times4 = 9600\); With A and 5: \(^4P_1\times\ ^6P_3\times8 = 3840\) | M1 | One outcome correct, accept unsimplified. |
| Required number \(= 6720 + 9600 - 3840\) | M1 | Adding 'with a' to 'with 5' and subtracting 'A and 5'. |
| \(12480\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((1\times)\ 3\times\ ^5P_2\) | M1 | \(3\times\ ^5P_n,\ n=2,3\). Condone \(3\times\ ^5C_2\), no \(+\) or \(-\). |
| Probability \(= \dfrac{\textit{their}\ 3\times5P2}{\textit{their}\ 16800}\) | M1 | Probability \(= \dfrac{\textit{their}\ 60}{\textit{their}\ 16800}\) |
| \(\dfrac{1}{280},\ 0.00357\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{1}{5}\times\dfrac{1}{4}\times\dfrac{1}{7}\times\dfrac{3}{6}\!\left(\dfrac{5}{5}\times\dfrac{4}{4}\right)\) or \(\dfrac{1}{5}\times\dfrac{1}{4}\times\dfrac{3\times5P2}{7P4}\) | M1 | \(\dfrac{1}{5}\times\dfrac{1}{4}\times k\) where \(0 < k < 1\) for considering letters. |
| M1 | \(t\times\dfrac{1}{7}\times\dfrac{3}{6}\) or \(t\times\dfrac{3\times5P2}{7P4}\) where \(0 < t < 1\). | |
| \(\dfrac{1}{280}\) | A1 | CAO |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $^5P_2 \times\ ^7P_4$ or $5\times4\times7\times6\times5\times4$ | M1 | $^5P_x \times\ ^7P_y,\ 1\leq x\leq 4,\ 1\leq y\leq 6$ |
| $16800$ | A1 | |
---
## Question 5(b):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| With A and no 5: $8\times\ ^6P_4$ or $(1\times4\times6\times5\times4\times3)\times2$ or $4C1\times2!\times6P4 = 2880$ | M1 | One number of ways correct, accept unsimplified. |
| With 5 and no A: $^4P_2\times4\times\ ^6P_3 = 5760$; With A and 5: $8\times4\times\ ^6P_3 = 3840$ | M1 | Add 2 or 3 identified correct scenarios only, accept unsimplified. |
| Total $= 12480$ | A1 | CAO |
**Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| No A or 5: $(4\times3)\times(6\times5\times4\times3) = 4320$ | M1 | $^4P_2 \times\ ^6P_4$ or $^4C_2\times\ ^6C_4$ seen, accept unsimplified. |
| Required number $= \textit{their}\ \textbf{(a)} - \textit{their}\ 4320$ | M1 | *Their* **5(a)** (or correct) $-$ *their* (No A or 5) value. |
| $12480$ | A1 | |
**Method 3:**
| Answer | Mark | Guidance |
|--------|------|----------|
| With A: $^4P_1\times\ ^7P_4\times2 = 6720$; With 5: $^5P_2\times\ ^6P_3\times4 = 9600$; With A and 5: $^4P_1\times\ ^6P_3\times8 = 3840$ | M1 | One outcome correct, accept unsimplified. |
| Required number $= 6720 + 9600 - 3840$ | M1 | Adding 'with a' to 'with 5' and subtracting 'A and 5'. |
| $12480$ | A1 | CAO |
---
## Question 5(c):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(1\times)\ 3\times\ ^5P_2$ | M1 | $3\times\ ^5P_n,\ n=2,3$. Condone $3\times\ ^5C_2$, no $+$ or $-$. |
| Probability $= \dfrac{\textit{their}\ 3\times5P2}{\textit{their}\ 16800}$ | M1 | Probability $= \dfrac{\textit{their}\ 60}{\textit{their}\ 16800}$ |
| $\dfrac{1}{280},\ 0.00357$ | A1 | |
**Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{1}{5}\times\dfrac{1}{4}\times\dfrac{1}{7}\times\dfrac{3}{6}\!\left(\dfrac{5}{5}\times\dfrac{4}{4}\right)$ or $\dfrac{1}{5}\times\dfrac{1}{4}\times\dfrac{3\times5P2}{7P4}$ | M1 | $\dfrac{1}{5}\times\dfrac{1}{4}\times k$ where $0 < k < 1$ for considering letters. |
| | M1 | $t\times\dfrac{1}{7}\times\dfrac{3}{6}$ or $t\times\dfrac{3\times5P2}{7P4}$ where $0 < t < 1$. |
| $\dfrac{1}{280}$ | A1 | CAO |
---5 A security code consists of 2 letters followed by a 4-digit number. The letters are chosen from $\{ \mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D } , \mathrm { E } \}$ and the digits are chosen from $\{ 1,2,3,4,5,6,7 \}$. No letter or digit may appear more than once. An example of a code is BE 3216 .
\begin{enumerate}[label=(\alph*)]
\item How many different codes can be formed?
\item Find the number of different codes that include the letter A or the digit 5 or both.\\
A security code is formed at random.
\item Find the probability that the code is DE followed by a number between 4500 and 5000 .
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q5 [8]}}