| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Selection with type constraints |
| Difficulty | Moderate -0.8 This is a straightforward permutations and combinations question testing basic counting principles. Part (i) requires simple reasoning about fixed positions, while parts (ii)-(iv) involve routine selection with simple constraints on repeated letters. All parts use standard techniques with no novel insight required, making it easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{7!}{3!} \times 2 = 1680\) | B1 B1 [2] | \(\frac{7!}{3!}\) or 840 seen or implied; correct answer |
| (ii) \(^6C_4 = 15\) | B1 [1] | correct answer |
| (iii) 1E in \(^6C_3\) ways \(= 20\) | M1 A1 [2] | \(k \times {^6C_x}\) or \(k \times {^6C_5}\) (\(k\) a constant) or \(^6P_4\) or \(^6P_3\) seen; correct final answer |
| (iv) need 2Es in \(^6C_2\) ways = 15 ways; need 3Es in \(^6C_1 = 6\) ways; total = 15 + 20 + 15 + 6 = 56 ways | M1 A1 M1 A1ft [4] | attempt to find ways with 2Es or 3Es; \(^6C_2\) oe and \(^6C_1\) oe seen; summing ways for no Es, 1E, 2Es and 3Es; correct final answer, ft on their four answers |
(i) $\frac{7!}{3!} \times 2 = 1680$ | B1 B1 [2] | $\frac{7!}{3!}$ or 840 seen or implied; correct answer
(ii) $^6C_4 = 15$ | B1 [1] | correct answer
(iii) 1E in $^6C_3$ ways $= 20$ | M1 A1 [2] | $k \times {^6C_x}$ or $k \times {^6C_5}$ ($k$ a constant) or $^6P_4$ or $^6P_3$ seen; correct final answer
(iv) need 2Es in $^6C_2$ ways = 15 ways; need 3Es in $^6C_1 = 6$ ways; total = 15 + 20 + 15 + 6 = 56 ways | M1 A1 M1 A1ft [4] | attempt to find ways with 2Es or 3Es; $^6C_2$ oe and $^6C_1$ oe seen; summing ways for no Es, 1E, 2Es and 3Es; correct final answer, ft on their four answers3 (i) In how many ways can all 9 letters of the word TELEPHONE be arranged in a line if the letters P and L must be at the ends?
How many different selections of 4 letters can be made from the 9 letters of the word TELEPHONE if\\
(ii) there are no Es,\\
(iii) there is exactly 1 E ,\\
(iv) there are no restrictions?
\hfill \mbox{\textit{CAIE S1 2012 Q3 [9]}}