| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Selection from categorized items |
| Difficulty | Moderate -0.8 This is a straightforward combinations question with standard restrictions. Part (i) is direct application of C(11,6), part (ii) requires simple case-by-case addition (4,5,6 from A), and part (iii) involves basic complementary counting. All techniques are routine for S1 level with no novel problem-solving required. |
| Spec | 5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(_{11}C_6 = 462\) | B1 | |
| OR \(A3 \, B3\) or \(A4 \, B2\) or \(A5 \, B1\) or \(A6\) | B1 | |
| \(= {}_3C_3 + {}_4C_4 \times {}_2C_2 + {}_5C_5 \times {}_3C_3 + {}_6C_6\) | B1 | |
| \(= 56 + 210 + 168 + 28 = 462\) | [1] | |
| (ii) \({}_{4}C_4 \times {}_2C_2 + {}_5C_5 \times {}_3C_3 + {}_6C_6\) | M1 | \(\Sigma\) 2 or more two-factor terms, \(P\) or \(C\) any numbers |
| \(= 210 + 168 + 28 = 406\) | B1 | Any correct option unsimplified |
| A1 [3] | Correct answer | |
| (iii) \({}_{4}C_4 + {}_6C_6 = 126 + 84\) | M1 | Summing \(_C_5 + {}_6C_6\) can be mult by 2 no other terms |
| \(= 210\) | B1 | 126 or 84 seen or unsimplified \({}_4C_4, {}_6C_6\) |
| A1 | Correct answer | |
| OR: 1,2 in A tog with: \(A1B3 + A2B2 + A3B1 + A4B0\), 1,2 out of A: \(A3B3 + A4B2 + A5B1 + A6B0\) | M1 | \(\Sigma\) 5 or more 2-factor \(_nP_r\) or \(_nC_r\) with \(_nC_r\) or \(_nP_r\) only (can be mult by 2) |
| \(= {}_1C_1 + {}_2C_1 \times {}_2C_2 + {}_3C_1 \times {}_3C_3 + {}_4C_1 \times {}_4C_4 + {}_3C_1 \times {}_3C_3 + {}_4C_4 \times {}_2C_2 + {}_5C_1 + {}_6C_6\) | B1 | 3 or more correct unsimplified options |
| \(= 6 + 45 + 60 + 15 + 20 + 45 + 18 + 1 = 210\) | A1 | |
| OR: \(462 - {}_6C_5 - {}_5C_5\) | M1 | subt two \(_nC_r\) options from their (i) \(_nC_r\) seen oe if using this method |
| \(= 210\) | A1 [3] | Correct answer |
(i) $_{11}C_6 = 462$ | B1 |
OR $A3 \, B3$ or $A4 \, B2$ or $A5 \, B1$ or $A6$ | B1 |
$= {}_3C_3 + {}_4C_4 \times {}_2C_2 + {}_5C_5 \times {}_3C_3 + {}_6C_6$ | B1 |
$= 56 + 210 + 168 + 28 = 462$ | [1] |
(ii) ${}_{4}C_4 \times {}_2C_2 + {}_5C_5 \times {}_3C_3 + {}_6C_6$ | M1 | $\Sigma$ 2 or more two-factor terms, $P$ or $C$ any numbers
$= 210 + 168 + 28 = 406$ | B1 | Any correct option unsimplified
| A1 [3] | Correct answer
(iii) ${}_{4}C_4 + {}_6C_6 = 126 + 84$ | M1 | Summing $_C_5 + {}_6C_6$ can be mult by 2 no other terms
$= 210$ | B1 | 126 or 84 seen or unsimplified ${}_4C_4, {}_6C_6$
| A1 | Correct answer
OR: 1,2 in A tog with: $A1B3 + A2B2 + A3B1 + A4B0$, 1,2 out of A: $A3B3 + A4B2 + A5B1 + A6B0$ | M1 | $\Sigma$ 5 or more 2-factor $_nP_r$ or $_nC_r$ with $_nC_r$ or $_nP_r$ only (can be mult by 2)
$= {}_1C_1 + {}_2C_1 \times {}_2C_2 + {}_3C_1 \times {}_3C_3 + {}_4C_1 \times {}_4C_4 + {}_3C_1 \times {}_3C_3 + {}_4C_4 \times {}_2C_2 + {}_5C_1 + {}_6C_6$ | B1 | 3 or more correct unsimplified options
$= 6 + 45 + 60 + 15 + 20 + 45 + 18 + 1 = 210$ | A1 |
OR: $462 - {}_6C_5 - {}_5C_5$ | M1 | subt two $_nC_r$ options from their (i) $_nC_r$ seen oe if using this method
$= 210$ | A1 [3] | Correct answer5 An English examination consists of 8 questions in Part $A$ and 3 questions in Part $B$. Candidates must choose 6 questions. The order in which questions are chosen does not matter. Find the number of ways in which the 6 questions can be chosen in each of the following cases.\\
(i) There are no restrictions on which questions can be chosen.\\
(ii) Candidates must choose at least 4 questions from Part $A$.\\
(iii) Candidates must either choose both question 1 and question 2 in Part $A$, or choose neither of these questions.
\hfill \mbox{\textit{CAIE S1 2012 Q5 [7]}}