| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Digit arrangements forming numbers |
| Difficulty | Standard +0.3 This is a standard permutations with repetitions problem requiring systematic case-by-case counting. Part (i) uses block arrangements and restricted positions with identical items (9!/(2!2!3!) with constraints). Part (ii) requires careful enumeration of cases for selecting and arranging cards with repetitions. While multi-part and requiring organization, the techniques are routine for A-level statistics/discrete maths with no novel insight needed—slightly easier than average due to being methodical rather than conceptually demanding. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6!\) | M1 | Seen in a single term expression as numerator |
| \((\times)\ 4!\) OR \((\times)\ 4 \times 3\) | M1 | Seen in a single term expression as numerator (denominator may be 1) |
| \(\div\ 2!2!3!\) OR \(\div\ 2!3!\) | M1 | Seen in a single term expression as denominator |
| Total \(= 720\) ways | A1 | Correct answer — Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1{*}{*}{*}{*}{*}{*}{*}3 = \dfrac{7!}{3!2!} = 420\) | B1 | \(\dfrac{7!}{3!2!}\) seen oe |
| \(3{*}{*}{*}{*}{*}{*}{*}1 = 420\) | M1 | Attempting to evaluate and sum at least 2 of \(1{*}{*}{*}3,\ 3{*}{*}{*}1,\ 3{*}{*}{*}3\) |
| \(3{*}{*}{*}{*}{*}{*}{*}3 = 420\) | ||
| Total \(= 1260\) ways | A1 | Correct answer — Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5 \times 4 \times 3 = 60\) ways \(({}^5P_3)\) | M1 | \({}^5P_3\) or \({}^5C_3 \times 3!\) (can be implied) |
| A1 | Correct answer — Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2^{**}\) in: 212, 213, 214, 216, 221, 223, 224, 226, 231, 232, 233, 234, 236, 241, 242, 243, 246, 261, 262, 263, 264, 266 | M1 | Listing attempt starting with 2, at least 10 correct entries |
| Total \(= 22\) ways | A1 | Correct answer — Total: 2 |
| Alternative: \(3 \times {}^4C_1 + 2 \times {}^5C_1\) | M1 | \(p \times {}^4C_1 + q \times {}^5C_1\), oe \(p+q > 2\) |
| OR \({}^5P_2 + {}^2C_1\) | M1 | \({}^5P_2\) seen |
| OR \({}^4P_2 + 2 \times {}^4P_1 + {}^2C_1\) | M1 | Any 2 terms added |
## Question 7:
### Part (i)(a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6!$ | M1 | Seen in a single term expression as numerator |
| $(\times)\ 4!$ **OR** $(\times)\ 4 \times 3$ | M1 | Seen in a single term expression as numerator (denominator may be 1) |
| $\div\ 2!2!3!$ **OR** $\div\ 2!3!$ | M1 | Seen in a single term expression as denominator |
| Total $= 720$ ways | A1 | Correct answer — **Total: 4** |
### Part (i)(b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1{*}{*}{*}{*}{*}{*}{*}3 = \dfrac{7!}{3!2!} = 420$ | B1 | $\dfrac{7!}{3!2!}$ seen oe |
| $3{*}{*}{*}{*}{*}{*}{*}1 = 420$ | M1 | Attempting to evaluate and sum at least 2 of $1{*}{*}{*}3,\ 3{*}{*}{*}1,\ 3{*}{*}{*}3$ |
| $3{*}{*}{*}{*}{*}{*}{*}3 = 420$ | | |
| Total $= 1260$ ways | A1 | Correct answer — **Total: 3** |
### Part (ii)(a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5 \times 4 \times 3 = 60$ ways $({}^5P_3)$ | M1 | ${}^5P_3$ or ${}^5C_3 \times 3!$ (can be implied) |
| | A1 | Correct answer — **Total: 2** |
### Part (ii)(b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2^{**}$ in: 212, 213, 214, 216, 221, 223, 224, 226, 231, 232, 233, 234, 236, 241, 242, 243, 246, 261, 262, 263, 264, 266 | M1 | Listing attempt starting with 2, at least 10 correct entries |
| Total $= 22$ ways | A1 | Correct answer — **Total: 2** |
| **Alternative:** $3 \times {}^4C_1 + 2 \times {}^5C_1$ | M1 | $p \times {}^4C_1 + q \times {}^5C_1$, oe $p+q > 2$ |
| **OR** ${}^5P_2 + {}^2C_1$ | M1 | ${}^5P_2$ seen |
| **OR** ${}^4P_2 + 2 \times {}^4P_1 + {}^2C_1$ | M1 | Any 2 terms added |7 Nine cards are numbered $1,2,2,3,3,4,6,6,6$.\\
(i) All nine cards are placed in a line, making a 9-digit number. Find how many different 9-digit numbers can be made in this way
\begin{enumerate}[label=(\alph*)]
\item if the even digits are all together,
\item if the first and last digits are both odd.\\
(ii) Three of the nine cards are chosen and placed in a line, making a 3-digit number. Find how many different numbers can be made in this way\\
(a) if there are no repeated digits,\\
(b) if the number is between 200 and 300 .
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2014 Q7 [11]}}