| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | At least/at most N letters between items |
| Difficulty | Standard +0.8 Part (a) is routine permutations with repetition. Part (b) requires careful casework to count arrangements with positional constraints (at least 5 letters between As), which is non-trivial. Part (c) involves combinatorial probability with multiple cases (comparing counts of As vs Ts in selections), requiring systematic enumeration. The multi-part structure with increasing complexity and the need for organized case analysis places this above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{9!}{2!2!}\) | M1 | \(\frac{h!}{2! \times j!}\), \(h = 7, 8, 9\); \(j = 1, 2\) |
| \(90720\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| With gap of 5: \(\frac{7!}{2!} \times 3\ [= 7560]\) | M1 | \(\frac{7!}{2!} \times k\), \(k\) positive integer \(1 < k < 7\) |
| With gap of 6: \(\frac{7!}{2!} \times 2\ [= 5040]\) | M1 | Add their no of ways for 3 identified correct scenarios, no additional incorrect scenarios, accept unsimplified |
| Answer | Marks |
|---|---|
| \(\left[\text{Total} = \frac{7!}{2!} \times 6 =\right] 15120\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(AT\text{---}: 2\times2\times{}^5C_3 = 40\); \(A\text{----}: 2\times{}^5C_4 = 10\); \(A\overline{A}T\overline{T}\text{-}: {}^5C_1 = 5\); \(AAT\text{--}: 2\times{}^5C_2 = 20\); \(AA\text{---}: {}^5C_3 = 10\); \(\text{-----}: {}^5C_5 = 1\) | B1 | Correct no of ways for 4 correctly identified scenarios, accept unsimplified |
| Total \(= 40+10+5+20+10+1\ [= 86]\) | M1 | Add no of ways for 5 or 6 identified correct scenarios, no additional incorrect scenarios, accept unsimplified |
| A1 | All correct and added | |
| \(\text{Probability} = \frac{86}{{}^9C_5}\) | M1 | \(\frac{\textit{their}\ 86}{{}^9C_5\ \text{or their identified total}}\), accept numerator unevaluated |
| \(\frac{86}{126}, \frac{43}{63},\ 0.683\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T\text{----}: 2\times{}^5C_4 = 10\); \(TTA\text{--}: 2\times{}^5C_2 = 20\); \(TT\text{---}: {}^5C_3 = 10\) | B1 | Correct no of ways for 2 correctly identified scenarios |
| M1 | Add no of ways for 2 or 3 correct scenarios and subtract from total; all correct and subtracted | |
| Total with more T's than A's \(= 40\); \({}^9C_5 - 40 = 86\) | A1 | |
| \(\text{Probability} = \frac{86}{{}^9C_5}\) | M1 | \(\frac{\textit{their}\ 86}{{}^9C_5}\), accept numerator unevaluated |
| \(\frac{43}{63},\ 0.683\) | A1 |
# Question 6(a):
$\frac{9!}{2!2!}$ | M1 | $\frac{h!}{2! \times j!}$, $h = 7, 8, 9$; $j = 1, 2$
$90720$ | A1 |
---
# Question 6(b):
With gap of 5: $\frac{7!}{2!} \times 3\ [= 7560]$ | M1 | $\frac{7!}{2!} \times k$, $k$ positive integer $1 < k < 7$
With gap of 6: $\frac{7!}{2!} \times 2\ [= 5040]$ | M1 | Add their no of ways for 3 identified correct scenarios, no additional incorrect scenarios, accept unsimplified
With gap of 7: $\frac{7!}{2!} \times 1\ [= 2520]$
$\left[\text{Total} = \frac{7!}{2!} \times 6 =\right] 15120$ | A1 |
---
# Question 6(c):
**Method 1:**
$AT\text{---}: 2\times2\times{}^5C_3 = 40$; $A\text{----}: 2\times{}^5C_4 = 10$; $A\overline{A}T\overline{T}\text{-}: {}^5C_1 = 5$; $AAT\text{--}: 2\times{}^5C_2 = 20$; $AA\text{---}: {}^5C_3 = 10$; $\text{-----}: {}^5C_5 = 1$ | B1 | Correct no of ways for 4 correctly identified scenarios, accept unsimplified
Total $= 40+10+5+20+10+1\ [= 86]$ | M1 | Add no of ways for 5 or 6 identified correct scenarios, no additional incorrect scenarios, accept unsimplified
| A1 | All correct and added
$\text{Probability} = \frac{86}{{}^9C_5}$ | M1 | $\frac{\textit{their}\ 86}{{}^9C_5\ \text{or their identified total}}$, accept numerator unevaluated
$\frac{86}{126}, \frac{43}{63},\ 0.683$ | A1 |
**Method 2:**
$T\text{----}: 2\times{}^5C_4 = 10$; $TTA\text{--}: 2\times{}^5C_2 = 20$; $TT\text{---}: {}^5C_3 = 10$ | B1 | Correct no of ways for 2 correctly identified scenarios
| M1 | Add no of ways for 2 or 3 correct scenarios and subtract from total; all correct and subtracted
Total with more T's than A's $= 40$; ${}^9C_5 - 40 = 86$ | A1 |
$\text{Probability} = \frac{86}{{}^9C_5}$ | M1 | $\frac{\textit{their}\ 86}{{}^9C_5}$, accept numerator unevaluated
$\frac{43}{63},\ 0.683$ | A1 |
---6
\begin{enumerate}[label=(\alph*)]
\item Find the number of different arrangements of the 9 letters in the word ACTIVATED.
\item Find the number of different arrangements of the 9 letters in the word ACTIVATED in which there are at least 5 letters between the two As.\\
Five letters are selected at random from the 9 letters in the word ACTIVATED.
\item Find the probability that the selection does not contain more Ts than As.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q6 [10]}}