| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with positional constraints |
| Difficulty | Moderate -0.3 This is a standard permutations question with repeated letters requiring systematic application of formulas. Parts (i)-(iii) use 11!/(repetitions) with straightforward restrictions (fixing positions, complement counting). Part (iv) involves basic selection with constraints. All techniques are routine for S1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks |
|---|---|
| (i) 1663200 | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Number of ways \(= \frac{9!}{3!2!} = 30240\) | M1, A1 [2] | 9! or 9P9 seen; Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 40320\) | M1, M1 | 8!/2!2! seen mult by something; 4 oe 4!/3! or 4C1 etc. seen mult by something |
| \(1663200 - 40320 = 1622880\) | B1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Total \(= 10\) ways | M1, B1, A1 [3] | Summing 2 options; One option correct; Correct answer |
| OR \(5C2 = 10\) | M2, A1 | M1 for k5C2; Correct ans |
**(i)** 1663200 | B1 [1] |
**(ii)** M xxxxxxxx M
Number of ways $= \frac{9!}{3!2!} = 30240$ | M1, A1 [2] | 9! or 9P9 seen; Correct answer
**(iii)** 4 vowels together $= 8! \div 4/2!2!$
$= 40320$ | M1, M1 | 8!/2!2! seen mult by something; 4 oe 4!/3! or 4C1 etc. seen mult by something
$1663200 - 40320 = 1622880$ | B1 [3] | Correct answer
**(iv)** Exactly 2 Es $4C2 = 6$
Exactly 3 Es $4C1 = 4$
Total $= 10$ ways | M1, B1, A1 [3] | Summing 2 options; One option correct; Correct answer
OR $5C2 = 10$ | M2, A1 | M1 for k5C2; Correct ans6 The 11 letters of the word REMEMBRANCE are arranged in a line.\\
(i) Find the number of different arrangements if there are no restrictions.\\
(ii) Find the number of different arrangements which start and finish with the letter M .\\
(iii) Find the number of different arrangements which do not have all 4 vowels ( $\mathrm { E } , \mathrm { E } , \mathrm { A } , \mathrm { E }$ ) next to each other.
4 letters from the letters of the word REMEMBRANCE are chosen.\\
(iv) Find the number of different selections which contain no Ms and no Rs and at least 2 Es.
\hfill \mbox{\textit{CAIE S1 2013 Q6 [9]}}