| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Multi-stage selection problems |
| Difficulty | Standard +0.8 This is a multi-stage combinatorics problem requiring careful case analysis in part (i) to enumerate valid selections under multiple constraints, followed by permutation problems in (ii) and (iii) involving grouping and arrangement restrictions. The case enumeration and constraint handling elevate this above routine combination exercises, though the individual calculations are standard A-level techniques. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M=3, R=1, O=2: \(= {}_7C3 \times {}_5C1 \times {}_8C2 = 4900\) | M1 | Summing more than one 3-term option involving combs (can be added) |
| M=3, R=2, O=1: \(= {}_7C3 \times {}_5C2 \times {}_8C1 = 2800\) | M1 | Mult 3 combs only (indep) |
| M=2, R=2, O=2: \(= {}_7C2 \times {}_5C2 \times {}_8C2 = 5880\) | A1 | 1 option correct unsimplified |
| Total \(= 13580\) | A1 [4] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 4 groups in \(4!\) ways; 3 mountain in \(3!\) ways; 2 ordinary in \(2!\) ways | M1 | \(4!\) seen mult by something |
| M1 | Mult by \(3!\) for racing or \(2!\) for ordinary | |
| \(4! \times 3! \times 2 = 288\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. s O x x x x O s s s; ordinary in \(2!\); rest of bikes in \(4!\); bikes and spaces 5 groups in 5 ways | M1 | \(2!\) or \(4!\) seen mult |
| M1 | Mult by 5 (ssssb) | |
| \(2! \times 4! \times 5 = 240\) | A1 [3] | Correct answer |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| M=3, R=1, O=2: $= {}_7C3 \times {}_5C1 \times {}_8C2 = 4900$ | M1 | Summing more than one 3-term option involving combs (can be added) |
| M=3, R=2, O=1: $= {}_7C3 \times {}_5C2 \times {}_8C1 = 2800$ | M1 | Mult 3 combs only (indep) |
| M=2, R=2, O=2: $= {}_7C2 \times {}_5C2 \times {}_8C2 = 5880$ | A1 | 1 option correct unsimplified |
| Total $= 13580$ | A1 **[4]** | Correct answer |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 4 groups in $4!$ ways; 3 mountain in $3!$ ways; 2 ordinary in $2!$ ways | M1 | $4!$ seen mult by something |
| | M1 | Mult by $3!$ for racing or $2!$ for ordinary |
| $4! \times 3! \times 2 = 288$ | A1 **[3]** | Correct answer |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. s O x x x x O s s s; ordinary in $2!$; rest of bikes in $4!$; bikes and spaces 5 groups in 5 ways | M1 | $2!$ or $4!$ seen mult |
| | M1 | Mult by 5 (ssssb) |
| $2! \times 4! \times 5 = 240$ | A1 **[3]** | Correct answer |
---6 A shop has 7 different mountain bicycles, 5 different racing bicycles and 8 different ordinary bicycles on display. A cycling club selects 6 of these 20 bicycles to buy.\\
(i) How many different selections can be made if there must be no more than 3 mountain bicycles and no more than 2 of each of the other types of bicycle?
The cycling club buys 3 mountain bicycles, 1 racing bicycle and 2 ordinary bicycles and parks them in a cycle rack, which has a row of 10 empty spaces.\\
(ii) How many different arrangements are there in the cycle rack if the mountain bicycles are all together with no spaces between them, the ordinary bicycles are both together with no spaces between them and the spaces are all together?\\
(iii) How many different arrangements are there in the cycle rack if the ordinary bicycles are at each end of the bicycles and there are no spaces between any of the bicycles?
\hfill \mbox{\textit{CAIE S1 2013 Q6 [10]}}