| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| 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 repeated letters requiring systematic application of formulas. Part (a) is routine (8!/3!2!2!), part (b) requires fixing positions then subtracting cases where Os are together (straightforward constraint handling), and part (c) involves complementary counting with combinations. All techniques are textbook exercises with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{8!}{2!\,3!}\) | M1 | \(\frac{8!}{k!\times m!}\), \(k = 1\) or \(2\), \(m = 1\) or \(3\), not \(k = m = 1\), no additional terms |
| \(3360\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{6!}{3!} - 4!\) | M1 | \(\frac{6!}{k!} - m\), \(1 \leq k \leq 3\), \(m\) an integer, condone \(2 \times \left(\frac{6!}{k!}\right) - m\) |
| \(w - 4!\) or \(w - 24\), \(w\) an integer | M1 | Condone \(w - 2 \times 4!\) |
| \(96\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(^4C_3 \times 3! + {^4C_2} \times 2 \times 3!\) | M1 | \(^4C_3 \times 3! + r\) or \(4 \times 3! + r\) or \(^4P_3 \times 3! + r\), \(r\) an integer. Condone \(2 \times {^4C_3} \times 3! + r\), \(2 \times 4 \times 3! + r\) or \(2 \times {^4P_3} \times 3! + r\) |
| M1 | \(q + {^4C_2} \times 3! \times k\) or \(q + {^4P_2} \times 3! \times k\), \(k = 1,2\), \(q\) an integer | |
| \([24 + 72 =] \ 96\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| OORR: \(^3C_2 \times {^2C_2} \times [^3C_0] = 3 \times 1 = 3\) | B1 | Outcomes for 2 identifiable scenarios correct, accept unsimplified |
| ORR\_: \(^3C_1 \times {^2C_2} \times {^3C_1} = 3 \times 1 \times 3 = 9\) | ||
| OOR\_: \(^3C_2 \times {^2C_1} \times {^3C_1} = 3 \times 2 \times 3 = 18\) | M1 | Add 4 or 5 identified correct scenarios only values, no additional incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations |
| OR\_\_: \(^3C_1 \times {^2C_1} \times {^3C_2} = 3 \times 2 \times 3 = 18\) | ||
| OOOR: \(^3C_3 \times {^2C_1} \times [^3C_0] = 1 \times 2 = 2\) | ||
| Total \(50\) | A1 | All correct and added |
| Probability \(= \dfrac{50}{^8C_4}\) | M1 | \(\dfrac{\textit{their } 50}{^8C_4}\), accept numerator unevaluated |
| \(\dfrac{50}{70}\) or \(0.714\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| ORTM, ORTW, ORMW: \(^3C_1 \times {^2C_1} = 6\) each | B1 | Outcomes for 5 identifiable scenarios correct, accept unsimplified |
| ORRM, ORRW, ORRT: \(^3C_1 \times {^2C_2} = 3\) each | ||
| OROR, OROT, OROM, OROW: \(^3C_2 \times {^2C_1} = 6\) or \(^3C_2 \times {^2C_2} = 3\) | M1 | Add 9, 10 or 11 identified correct scenarios only values, no additional incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations |
| OROO: \(^3C_3 \times {^2C_1} = 2\) | ||
| Total \(50\) | A1 | All correct and added |
| Probability \(= \dfrac{50}{^8C_4}\) | M1 | \(\dfrac{\textit{their } 50}{^8C_4}\), accept numerator unevaluated |
| \(\dfrac{50}{70}\) or \(0.714\) | A1 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{8!}{2!\,3!}$ | M1 | $\frac{8!}{k!\times m!}$, $k = 1$ or $2$, $m = 1$ or $3$, not $k = m = 1$, no additional terms |
| $3360$ | A1 | |
## Question 6(b):
**Method 1:**
$\frac{6!}{3!} - 4!$ | M1 | $\frac{6!}{k!} - m$, $1 \leq k \leq 3$, $m$ an integer, condone $2 \times \left(\frac{6!}{k!}\right) - m$
$w - 4!$ or $w - 24$, $w$ an integer | M1 | Condone $w - 2 \times 4!$
$96$ | A1 |
**Method 2:**
$^4C_3 \times 3! + {^4C_2} \times 2 \times 3!$ | M1 | $^4C_3 \times 3! + r$ or $4 \times 3! + r$ or $^4P_3 \times 3! + r$, $r$ an integer. Condone $2 \times {^4C_3} \times 3! + r$, $2 \times 4 \times 3! + r$ or $2 \times {^4P_3} \times 3! + r$
| M1 | $q + {^4C_2} \times 3! \times k$ or $q + {^4P_2} \times 3! \times k$, $k = 1,2$, $q$ an integer
$[24 + 72 =] \ 96$ | A1 |
**Total: 3 marks**
---
## Question 6(c):
**Method 1:**
OORR: $^3C_2 \times {^2C_2} \times [^3C_0] = 3 \times 1 = 3$ | B1 | Outcomes for 2 identifiable scenarios correct, accept unsimplified
ORR\_: $^3C_1 \times {^2C_2} \times {^3C_1} = 3 \times 1 \times 3 = 9$ | |
OOR\_: $^3C_2 \times {^2C_1} \times {^3C_1} = 3 \times 2 \times 3 = 18$ | M1 | Add 4 or 5 identified correct scenarios only values, no additional incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations
OR\_\_: $^3C_1 \times {^2C_1} \times {^3C_2} = 3 \times 2 \times 3 = 18$ | |
OOOR: $^3C_3 \times {^2C_1} \times [^3C_0] = 1 \times 2 = 2$ | |
Total $50$ | A1 | All correct and added
Probability $= \dfrac{50}{^8C_4}$ | M1 | $\dfrac{\textit{their } 50}{^8C_4}$, accept numerator unevaluated
$\dfrac{50}{70}$ or $0.714$ | A1 |
**Method 2:**
ORTM, ORTW, ORMW: $^3C_1 \times {^2C_1} = 6$ each | B1 | Outcomes for 5 identifiable scenarios correct, accept unsimplified
ORRM, ORRW, ORRT: $^3C_1 \times {^2C_2} = 3$ each | |
OROR, OROT, OROM, OROW: $^3C_2 \times {^2C_1} = 6$ or $^3C_2 \times {^2C_2} = 3$ | M1 | Add 9, 10 or 11 identified correct scenarios only values, no additional incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations
OROO: $^3C_3 \times {^2C_1} = 2$ | |
Total $50$ | A1 | All correct and added
Probability $= \dfrac{50}{^8C_4}$ | M1 | $\dfrac{\textit{their } 50}{^8C_4}$, accept numerator unevaluated
$\dfrac{50}{70}$ or $0.714$ | A1 |
**Total: 5 marks**
---6
\begin{enumerate}[label=(\alph*)]
\item Find the total number of different arrangements of the 8 letters in the word TOMORROW.
\item Find the total number of different arrangements of the 8 letters in the word TOMORROW that have an R at the beginning and an R at the end, and in which the three Os are not all together.\\
Four letters are selected at random from the 8 letters of the word TOMORROW.
\item Find the probability that the selection contains at least one O and at least one R .
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q6 [10]}}