| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with grouped categories |
| Difficulty | Moderate -0.3 Part (a) is a standard 'treat grouped items as one unit' permutations problem with repeated letters, requiring careful counting of arrangements within and outside the vowel block. Part (b) is a straightforward complementary counting problem using combinations with constraints. Both are typical S1 exercises requiring methodical application of standard techniques rather than problem-solving insight, making this slightly easier than average for A-level. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. \(\)(AAOOOI)\(***\) | B1 | \(8!\ (8\times 7!)\) or \(6!\) seen anywhere, either alone or in numerator |
| \(\frac{8!}{2!2!}\times\frac{6!}{2!3!} = 604800\) | M1 | Dividing by at least 3 of \(2!2!2!3!\) (may be fractions added) |
| A1 (3) | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| C(7) E(6) A(4): \(1\ 1\ 2 = 7\times{}^6C_2\times 4 = 252\); \(1\ 2\ 1 = 7\times{}^6C_2\times 4 = 420\); \(1\ 3\ 0 = 7\times{}^6C_3\times 1 = 140\); \(2\ 1\ 1 = {}^7C_2\times 6\times 4 = 504\); \(2\ 2\ 0 = {}^7C_2\times{}^6C_2\times 1 = 315\); \(3\ 1\ 0 = {}^7C_3\times 6\times 1 = 210\) | M1 | Mult 3 appropriate combinations together; assume \(6={}^6C_1, 1={}^4C_0\) etc., \(\sum r=4\), C&E both present |
| A1 | At least 3 correct unsimplified products | |
| M1* DM1 | Listing at least 4 different correct options; summing at least 4 outcomes, involving 3 combs or perms, \(\sum r=4\) | |
| Total \(= 1841\) | A1 (5) | Correct answer |
| SC if CE removed, M1 available for listing at least 4 different correct options for remaining 2. DM1 for \({}^7C_1\times{}^6C_1\times\)(sum of at least 4 outcomes) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. $**$(AAOOOI)$*****$ | B1 | $8!\ (8\times 7!)$ or $6!$ seen anywhere, either alone or in numerator |
| $\frac{8!}{2!2!}\times\frac{6!}{2!3!} = 604800$ | M1 | Dividing by at least 3 of $2!2!2!3!$ (may be fractions added) |
| | A1 (3) | Correct answer |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| C(7) E(6) A(4): $1\ 1\ 2 = 7\times{}^6C_2\times 4 = 252$; $1\ 2\ 1 = 7\times{}^6C_2\times 4 = 420$; $1\ 3\ 0 = 7\times{}^6C_3\times 1 = 140$; $2\ 1\ 1 = {}^7C_2\times 6\times 4 = 504$; $2\ 2\ 0 = {}^7C_2\times{}^6C_2\times 1 = 315$; $3\ 1\ 0 = {}^7C_3\times 6\times 1 = 210$ | M1 | Mult 3 appropriate combinations together; assume $6={}^6C_1, 1={}^4C_0$ etc., $\sum r=4$, C&E both present |
| | A1 | At least 3 correct unsimplified products |
| | M1* DM1 | Listing at least 4 different correct options; summing at least 4 outcomes, involving 3 combs or perms, $\sum r=4$ |
| Total $= 1841$ | A1 (5) | Correct answer |
| | | SC if CE removed, M1 available for listing at least 4 different correct options for remaining 2. DM1 for ${}^7C_1\times{}^6C_1\times$(sum of at least 4 outcomes) |
---5
\begin{enumerate}[label=(\alph*)]
\item Find the number of different ways that the 13 letters of the word ACCOMMODATION can be arranged in a line if all the vowels (A, I, O) are next to each other.
\item There are 7 Chinese, 6 European and 4 American students at an international conference. Four of the students are to be chosen to take part in a television broadcast. Find the number of different ways the students can be chosen if at least one Chinese and at least one European student are included.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2015 Q5 [8]}}