| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Digit arrangements forming numbers |
| Difficulty | Moderate -0.8 This is a straightforward permutations and combinations question testing basic counting principles. Parts (i)(a) and (i)(b) are standard textbook exercises requiring simple multiplication (3! and 3³). Parts (i)(c) and (ii) require slightly more careful case analysis but use only elementary counting methods with no complex reasoning or novel insight required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks |
|---|---|
| 6 | B1, [1] |
| Answer | Marks | Guidance |
|---|---|---|
| \(3×3×3 = 27\); \(3! + 7×3\) or \(3 + 3×6 + 6\) or \(3! × 4 + 3\) | M1, A1, [2] | Complete correct method. Allow methods equiv to these. Only allow other methods if they appear correct |
| Answer | Marks | Guidance |
|---|---|---|
| (i)(b) − 3; If answer is not 24, this method must be explicitly stated in order to give M1A1ft | M1, A1fr, [2] | or 3! + 6×3 or 3! + 3!×3 or 6 + 3!×3!+2! or 3! × 4; Complete correct method. Allow methods equiv to these. Only allow other methods if they appear correct |
| Answer | Marks | Guidance |
|---|---|---|
| eg 1123: \(\frac{4!}{2!}\) × 3 alone; allow M1 for \(\frac{4!}{2!}\) × 3! alone; eg 1122: \(\frac{4!}{2!2!}\) allow M1 for \(\frac{4!}{2!}\) × 3! alone | M2, M1, [5] | \(3! × ^3C_1 × 3\) or \(3! × 12\) ÷ 2; M1dep (= 36); \(3! × ^3C_2\) or \(3!\) × \(\frac{4!}{2!}\) ÷ 2; M1dep (= 18); Allow methods equiv to these, eg correctly listing cases. Only allow other methods if they appear correct. NB \(3×3×2×2 = 36\) & \(3×3×2 × 1 = 18\) are incorrect methods unless clear justification given |
| Total = 54 | A1 | Correct ans, no working: M1M1A1 |
## (i)(a)
6 | B1, [1]
## (i)(b)
$3×3×3 = 27$; $3! + 7×3$ or $3 + 3×6 + 6$ or $3! × 4 + 3$ | M1, A1, [2] | Complete correct method. Allow methods equiv to these. Only allow other methods if they appear correct | (Explanation for 3! × 4 + 3: 123: 3!, 112 & 122: 3!, 223 & 233: 3!, 331 & 311: 3!; 111, 222, 333: 3. Candidates need not include this)
## (i)(c)
(i)(b) − 3; If answer is not 24, this method must be explicitly stated in order to give M1A1ft | M1, A1fr, [2] | or 3! + 6×3 or 3! + 3!×3 or 6 + 3!×3!+2! or 3! × 4; Complete correct method. Allow methods equiv to these. Only allow other methods if they appear correct | or 8 × 3 (Explanation: there are 8 possible orders starting with 1. Candidates need not include this)
## (ii)
eg 1123: $\frac{4!}{2!}$ × 3 alone; allow M1 for $\frac{4!}{2!}$ × 3! alone; eg 1122: $\frac{4!}{2!2!}$ allow M1 for $\frac{4!}{2!}$ × 3! alone | M2, M1, [5] | $3! × ^3C_1 × 3$ or $3! × 12$ ÷ 2; M1dep (= 36); $3! × ^3C_2$ or $3!$ × $\frac{4!}{2!}$ ÷ 2; M1dep (= 18); Allow methods equiv to these, eg correctly listing cases. Only allow other methods if they appear correct. NB $3×3×2×2 = 36$ & $3×3×2 × 1 = 18$ are incorrect methods unless clear justification given | This method only scores if $3x3x3x3 = \_...$: is used: M1; No. with 4 rep'ns = 3; M1; No. with 3 rep'ns = $\frac{4!}{3!}$; M1; No. with rep'ns = $\frac{4!}{3!}$ × 6 (= 24); M1; or 8 × 3; M2 (allow 81−(3'+24') or 81−27 M1 (allow 81−3 or 81−24)
Total = 54 | A1 | Correct ans, no working: M1M1A1 | 18, 36 only score if a correct method seen., or eg: 18 orders listed starting with ''1'' or 18 orders listed with two repetitions
---\begin{enumerate}[label=(\roman*)]
\item How many different 3-digit numbers can be formed using the digits 1, 2 and 3 when
\begin{enumerate}[label=(\alph*)]
\item no repetitions are allowed, [1]
\item any repetitions are allowed, [2]
\item each digit may be included at most twice? [2]
\end{enumerate}
\item How many different 4-digit numbers can be formed using the digits 1, 2 and 3 when each digit may be included at most twice? [5]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2013 Q4 [10]}}