| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Items NOT together (general separation) |
| Difficulty | Moderate -0.3 This is a straightforward permutations question with standard techniques: (a) counting arrangements with repeated letters, (b) treating a block as one unit, (c) using complement probability for 'not together'. All are textbook methods requiring minimal problem-solving insight, though the multi-part structure and careful counting place it slightly below average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[\frac{8!}{3!}\right] = 6720\) | B1 | NFWW, must be evaluated |
| Answer | Marks | Guidance |
|---|---|---|
| \(\_\_\_\text{LED}\_\_\) : With LED together: \(\frac{6!}{2!}\) | M1 | \(\frac{6!}{k}\) or \(\frac{5!x6}{k}\), \(k \geq 1\) and no other terms |
| \(\frac{m}{2!}\), \(m\) an integer, \(m \geq 5\) | M1 | \(\frac{m}{2!}\), \(m\) an integer, \(m \geq 5\) |
| \(360\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Arrange 6 letters RELESE \(= \frac{6!}{3!} [= 120]\) | *M1 | \(\frac{6!}{3!} \times k\) seen, \(k\) an integer \(> 0\) |
| Multiply by number of ways of placing AD in non-adjacent places \(= their\ 120 \times {}^7P_2 [= 5040]\) | *M1 | \(m \times n(n-1)\) or \(m \times {}^nC_2\) or \(m \times {}^nP_2\), \(n = 6, 7\) or \(8\), \(m\) an integer \(> 0\) |
| \([\text{Probability} =] \frac{their\ 5040}{their\ 6720}\) | DM1 | Denominator \(= their\) (a) or correct, dependent on at least one M mark already gained |
| \(\frac{5040}{6720}\) or \(\frac{3}{4}\) or \(0.75\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(their\ 6720 - \frac{7! \times 2}{3!} [= 5040]\) | *M1 | \(Their\ 6720 - k\), \(k\) a positive integer; \((m-)\frac{7! \times k}{3!}, k = 1,2\) |
| \([\text{Probability} =] \frac{their\ 5040}{their\ 6720}\) | DM1 | With denominator \(= their\) (a) or correct, dependent on at least one M mark already gained |
| \(\frac{5040}{6720}\) or \(\frac{3}{4}\) or \(0.75\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{7! \times 2}{3!} [= 1680]\) | *M1 | \(\frac{7! \times k}{3!}, k = 1,2\) |
| \([\text{Probability} =] \frac{their\ 1680}{their\ 6720}\) | *M1 | With denominator \(= their\) (a) or correct |
| \(1 - \frac{their\ 1680}{their\ 6720}\) | DM1 | \(1 - m\), \(0 < m < 1\), dependent on at least one M mark already gained |
| \(\frac{5040}{6720}\) or \(\frac{3}{4}\) or \(0.75\) | A1 |
## Question 3(a):
$\left[\frac{8!}{3!}\right] = 6720$ | B1 | NFWW, must be evaluated
---
## Question 3(b):
$\_\_\_\text{LED}\_\_$ : With LED together: $\frac{6!}{2!}$ | M1 | $\frac{6!}{k}$ or $\frac{5!x6}{k}$, $k \geq 1$ and no other terms
$\frac{m}{2!}$, $m$ an integer, $m \geq 5$ | M1 | $\frac{m}{2!}$, $m$ an integer, $m \geq 5$
$360$ | A1 | CAO
---
## Question 3(c):
**Method 1:**
Arrange 6 letters RELESE $= \frac{6!}{3!} [= 120]$ | *M1 | $\frac{6!}{3!} \times k$ seen, $k$ an integer $> 0$
Multiply by number of ways of placing AD in non-adjacent places $= their\ 120 \times {}^7P_2 [= 5040]$ | *M1 | $m \times n(n-1)$ or $m \times {}^nC_2$ or $m \times {}^nP_2$, $n = 6, 7$ or $8$, $m$ an integer $> 0$
$[\text{Probability} =] \frac{their\ 5040}{their\ 6720}$ | DM1 | Denominator $= their$ **(a)** or correct, dependent on at least one M mark already gained
$\frac{5040}{6720}$ or $\frac{3}{4}$ or $0.75$ | A1 |
**Alternative Method:**
$their\ 6720 - \frac{7! \times 2}{3!} [= 5040]$ | *M1 | $Their\ 6720 - k$, $k$ a positive integer; $(m-)\frac{7! \times k}{3!}, k = 1,2$
$[\text{Probability} =] \frac{their\ 5040}{their\ 6720}$ | DM1 | With denominator $= their$ **(a)** or correct, dependent on at least one M mark already gained
$\frac{5040}{6720}$ or $\frac{3}{4}$ or $0.75$ | A1 |
**Alternative Method 2 (using probability directly):**
$\frac{7! \times 2}{3!} [= 1680]$ | *M1 | $\frac{7! \times k}{3!}, k = 1,2$
$[\text{Probability} =] \frac{their\ 1680}{their\ 6720}$ | *M1 | With denominator $= their$ **(a)** or correct
$1 - \frac{their\ 1680}{their\ 6720}$ | DM1 | $1 - m$, $0 < m < 1$, dependent on at least one M mark already gained
$\frac{5040}{6720}$ or $\frac{3}{4}$ or $0.75$ | A1 |
---3
\begin{enumerate}[label=(\alph*)]
\item How many different arrangements are there of the 8 letters in the word RELEASED?
\item How many different arrangements are there of the 8 letters in the word RELEASED in which the letters LED appear together in that order?
\item An arrangement of the 8 letters in the word RELEASED is chosen at random. Find the probability that the letters A and D are not together.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q3 [8]}}