| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | November |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with identical objects |
| Difficulty | Easy -1.2 This is a straightforward application of basic counting principles: (i) uses the multiplication principle (2^12) and (ii) uses the standard formula for arrangements with identical objects (12!/(7!5!)). Both are direct recall of fundamental formulas with no problem-solving or insight required, making this easier than average for A-level. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| (i) each in 2 ways = \(2^{12} = 4096\) | M1, A1 | [2] \(2^{12}\) seen; Correct answer |
| (ii) \(\frac{12!}{7! 5!} = 792\) | B1 | [1] |
**(i)** each in 2 ways = $2^{12} = 4096$ | M1, A1 | [2] $2^{12}$ seen; Correct answer
**(ii)** $\frac{12!}{7! 5!} = 792$ | B1 | [1]2 Twelve coins are tossed and placed in a line. Each coin can show either a head or a tail.\\
(i) Find the number of different arrangements of heads and tails which can be obtained.\\
(ii) Find the number of different arrangements which contain 7 heads and 5 tails.
\hfill \mbox{\textit{CAIE S1 2011 Q2 [3]}}