| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with positional constraints |
| Difficulty | Standard +0.3 This is a standard permutations question with straightforward constraints. Part (a) uses the 'treat as one unit' technique, (b) requires fixing positions then subtracting unwanted cases, and (c) involves basic combinatorics with 'at least one' reasoning. All are textbook techniques for S1 level with no novel insight required, making it slightly easier than average A-level difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{8!}{2!2!}\) | M1 | \(\frac{k!}{2!m!}\), \(k = 7\) or \(8\), \(m = 1, 2\). |
| \(= 10080\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{7!}{2!2!} - \frac{6!}{2!}\) | M1 | \(\frac{7!}{2!2!} - r\), \(r\) integer \(> 1\). |
| \([= 1260 - 360]\) | M1 | \(s - \frac{6!}{2!}\), \(s\) integer \(> 360\). |
| \(= 900\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{5!}{2!} \times \frac{6\times5}{2}\) or \(\frac{5!}{2!} \times {}^6C_2\) | M1 | \(t \times \frac{6\times5}{2}\) or \(t \times {}^6C_2\), \(t\) an integer \(> 1\). |
| \([= 60 \times 15]\) | M1 | \(\frac{5!}{2!} \times u\), \(u\) an integer \(> 1\). |
| \(= 900\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| T E \(_{--}\) \(= {}^2C_1 \times {}^2C_1 \times {}^5C_2 = 40\) | B1 | Either identified or correct unsimplified expression, either alone or in an addition. |
| T E E \(_{-}\) \(= {}^2C_1 \times {}^2C_2 \times {}^5C_1 = 10\) | B1 | Either identified or correct unsimplified expression, either alone or in an addition. |
| Probability \(\frac{(40+10)}{{}^9C_4}\) | M1 | \(\frac{a}{{}^9C_4}\), \(a\) an integer \(< 126\). Denominator value must be seen as \({}^9C_4\) somewhere. |
| \(\left[\text{Percentage} = \frac{50}{126} \times 100\right] = 39.7\%\) | A1 | \(39.68 \leqslant \text{percentage} \leqslant 39.7\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| T \(\wedge\wedge\wedge\) \(= {}^2C_1 \times {}^7C_3 = 70\) | B1 | Either identified or correct unsimplified expression, either alone or in a subtraction. |
| T \(***\) \(= {}^2C_1 \times {}^5C_3 = 20\) | B1 | Either identified or correct unsimplified expression, either alone or in a subtraction. |
| Probability \(\frac{(70-20)}{{}^9C_4}\) | M1 | \(\frac{a}{{}^9C_4}\), \(a\) an integer \(< 126\). Denominator value must be seen as \({}^9C_4\) somewhere. |
| \(\left[\text{Percentage} = \frac{50}{126} \times 100\right] = 39.7\%\) | A1 | \(39.68 \leqslant \text{percentage} \leqslant 39.7\). |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{8!}{2!2!}$ | M1 | $\frac{k!}{2!m!}$, $k = 7$ or $8$, $m = 1, 2$. |
| $= 10080$ | A1 | |
---
## Question 7(b):
**Method 1:** Number of ways with no restriction on Es $-$ ways with Es together
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{7!}{2!2!} - \frac{6!}{2!}$ | M1 | $\frac{7!}{2!2!} - r$, $r$ integer $> 1$. |
| $[= 1260 - 360]$ | M1 | $s - \frac{6!}{2!}$, $s$ integer $> 360$. |
| $= 900$ | A1 | |
**Method 2:** T $\wedge\wedge\wedge\wedge\wedge$ T with Es inserted in gaps
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{5!}{2!} \times \frac{6\times5}{2}$ or $\frac{5!}{2!} \times {}^6C_2$ | M1 | $t \times \frac{6\times5}{2}$ or $t \times {}^6C_2$, $t$ an integer $> 1$. |
| $[= 60 \times 15]$ | M1 | $\frac{5!}{2!} \times u$, $u$ an integer $> 1$. |
| $= 900$ | A1 | |
---
## Question 7(c):
**Method 1 – addition:**
| Answer | Mark | Guidance |
|--------|------|----------|
| T E $_{--}$ $= {}^2C_1 \times {}^2C_1 \times {}^5C_2 = 40$ | B1 | Either identified or correct unsimplified expression, either alone or in an addition. |
| T E E $_{-}$ $= {}^2C_1 \times {}^2C_2 \times {}^5C_1 = 10$ | B1 | Either identified or correct unsimplified expression, either alone or in an addition. |
| Probability $\frac{(40+10)}{{}^9C_4}$ | M1 | $\frac{a}{{}^9C_4}$, $a$ an integer $< 126$. Denominator value must be seen as ${}^9C_4$ somewhere. |
| $\left[\text{Percentage} = \frac{50}{126} \times 100\right] = 39.7\%$ | A1 | $39.68 \leqslant \text{percentage} \leqslant 39.7$. |
**Method 2 – subtraction (total arrangements with 1 T $-$ number of arrangements with 1T 0 E):**
| Answer | Mark | Guidance |
|--------|------|----------|
| T $\wedge\wedge\wedge$ $= {}^2C_1 \times {}^7C_3 = 70$ | B1 | Either identified or correct unsimplified expression, either alone or in a subtraction. |
| T $***$ $= {}^2C_1 \times {}^5C_3 = 20$ | B1 | Either identified or correct unsimplified expression, either alone or in a subtraction. |
| Probability $\frac{(70-20)}{{}^9C_4}$ | M1 | $\frac{a}{{}^9C_4}$, $a$ an integer $< 126$. Denominator value must be seen as ${}^9C_4$ somewhere. |
| $\left[\text{Percentage} = \frac{50}{126} \times 100\right] = 39.7\%$ | A1 | $39.68 \leqslant \text{percentage} \leqslant 39.7$. |7
\begin{enumerate}[label=(\alph*)]
\item How many different arrangements are there of the 9 letters in the word INTELLECT in which the two Ts are together?
\item How many different arrangements are there of the 9 letters in the word INTELLECT in which there is a T at each end and the two Es are not next to each other?\\
Four letters are selected at random from the 9 letters in the word INTELLECT.\\[0pt]
\item Find the percentage of the possible selections which contain at least one E and exactly one T. [4]\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad3a6a8a-23fe-415a-b2f4-7c49136ccc6c-14_2715_31_106_2016}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q7 [9]}}