| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with divisibility constraints |
| Difficulty | Moderate -0.3 Part (a) is a standard permutations with repetition problem (9!/(2!×2!×3!×2!) and adjusting for even numbers), which is routine bookwork. Part (b) requires casework for 'at least one of each type' but is a straightforward application of combinations with simple enumeration of cases (1,1,3), (1,2,2), (2,1,2). Both parts are typical S1 exercises requiring methodical application of standard techniques rather than problem-solving insight, making this slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{9!}{2! \cdot 2! \cdot 3!} = 15120\) ways | B1 | Dividing by \(2! \cdot 2! \cdot 3!\) |
| B1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| ending in 3: \(\frac{8!}{2! \cdot 2! \cdot 3!} = 1680\) ways | B1 | Correct ways end in 3 |
| ending in 7: \(\frac{8!}{2! \cdot 3!} = 3360\) ways | B1 | Correct ways end in 7 |
| Total even \(= 15120 - 1680 - 3360\) | M1 | Finding odd and subtracting from 15120 or their (i) |
| \(= 10080\) ways | A1 [4] | Correct answer |
| OR | ||
| ending in 2: \(\frac{8!}{2! \cdot 3!} = 3360\) ways | B1 | One correct way end in even |
| ending in 6: \(\frac{8!}{2! \cdot 2! \cdot 3!} = 1680\) ways | B1 | Correct way end in another even |
| ending in 8: \(\frac{8!}{2! \cdot 2! \cdot 2!} = 5040\) ways | M1 | Summing 2 or 3 ways |
| Total \(= 10080\) ways | A1 | Correct answer |
| OR: \(15120 \times \frac{6}{9} = 10080\) | M2, A2 | Multiply their (i) by \(\frac{2}{3}\) oe; Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| T(3) S(6) G(14): | ||
| \(1\quad 1\): \(3 \times 6 \times {}^{14}C_3 = 6552\) | M1 | Multiply 3 (combinations) together; assume \(6 = {}^6C_1\) etc |
| \(1\quad 3\): \(3 \times {}^6C_3 \times 14 = 840\) | M1 | Listing at least 4 different options |
| \(3\quad 1\): \(1 \times 6 \times 14 = 84\) | M1 | Summing at least 4 different options |
| \(2\quad 2\): \({}^3C_2 \times {}^6C_2 \times 14 = 630\) | B1 | At least 3 correct numerical options |
| \(2\quad 1\): \({}^3C_2 \times 6 \times {}^{14}C_2 = 1638\) | ||
| \(1\quad 2\): \(3 \times {}^6C_2 \times {}^{14}C_2 = 4095\) | ||
| Total ways \(= 13839\) (13800) | A1 [5] | Correct answer |
## Question 7:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{9!}{2! \cdot 2! \cdot 3!} = 15120$ ways | B1 | Dividing by $2! \cdot 2! \cdot 3!$ |
| | B1 [2] | Correct answer |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| ending in 3: $\frac{8!}{2! \cdot 2! \cdot 3!} = 1680$ ways | B1 | Correct ways end in 3 |
| ending in 7: $\frac{8!}{2! \cdot 3!} = 3360$ ways | B1 | Correct ways end in 7 |
| Total even $= 15120 - 1680 - 3360$ | M1 | Finding odd and subtracting from 15120 or their (i) |
| $= 10080$ ways | A1 [4] | Correct answer |
| **OR** | | |
| ending in 2: $\frac{8!}{2! \cdot 3!} = 3360$ ways | B1 | One correct way end in even |
| ending in 6: $\frac{8!}{2! \cdot 2! \cdot 3!} = 1680$ ways | B1 | Correct way end in another even |
| ending in 8: $\frac{8!}{2! \cdot 2! \cdot 2!} = 5040$ ways | M1 | Summing 2 or 3 ways |
| Total $= 10080$ ways | A1 | Correct answer |
| **OR**: $15120 \times \frac{6}{9} = 10080$ | M2, A2 | Multiply their (i) by $\frac{2}{3}$ oe; Correct answer |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| T(3) S(6) G(14): | | |
| $1\quad 1$: $3 \times 6 \times {}^{14}C_3 = 6552$ | M1 | Multiply 3 (combinations) together; assume $6 = {}^6C_1$ etc |
| $1\quad 3$: $3 \times {}^6C_3 \times 14 = 840$ | M1 | Listing at least 4 different options |
| $3\quad 1$: $1 \times 6 \times 14 = 84$ | M1 | Summing at least 4 different options |
| $2\quad 2$: ${}^3C_2 \times {}^6C_2 \times 14 = 630$ | B1 | At least 3 correct numerical options |
| $2\quad 1$: ${}^3C_2 \times 6 \times {}^{14}C_2 = 1638$ | | |
| $1\quad 2$: $3 \times {}^6C_2 \times {}^{14}C_2 = 4095$ | | |
| Total ways $= 13839$ (13800) | A1 [5] | Correct answer |7
\begin{enumerate}[label=(\alph*)]
\item Find how many different numbers can be made by arranging all nine digits of the number 223677888 if
\begin{enumerate}[label=(\roman*)]
\item there are no restrictions,
\item the number made is an even number.
\end{enumerate}\item Sandra wishes to buy some applications (apps) for her smartphone but she only has enough money for 5 apps in total. There are 3 train apps, 6 social network apps and 14 games apps available. Sandra wants to have at least 1 of each type of app. Find the number of different possible selections of 5 apps that Sandra can choose.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2015 Q7 [11]}}