| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | People arrangements in groups/rows |
| Difficulty | Moderate -0.8 This is a straightforward permutations and combinations question with standard techniques. Part (i) is direct P(15,5), part (ii) applies the multiplication principle with given constraints, parts (iii-iv) require systematic case-by-case counting of combinations. All methods are routine textbook exercises requiring no novel insight, though the multi-part structure and careful counting make it slightly more substantial than trivial recall questions. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| \(^{15}P_5 = 360360\) | M1, A1 (2) | oe, can be implied; not \(^{15}C_5\); correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(5 \times 10 \times 4 \times 9 \times 3 = 5400\) | M1, A1 (2) | Mult 5 numbers; correct answer |
| Answer | Marks |
|---|---|
| M1, M1, A1 (3) | Mult 2 combs \(^5C_x \times {^{10}C_y}\); summing 2 or 3 two-factor options, \(x+y=5\); correct answer |
| Answer | Marks |
|---|---|
| M1, M1, A1 (3) | Mult 2 combs \(^4C_x\) and \(^9C_y\); summing both options \(x+y=3\), gender correct; correct answer |
## Question 6:
**(i)**
$^{15}P_5 = 360360$ | M1, A1 (2) | oe, can be implied; not $^{15}C_5$; correct answer
**(ii)**
$5 \times 10 \times 4 \times 9 \times 3 = 5400$ | M1, A1 (2) | Mult 5 numbers; correct answer
**(iii)**
M(5) F(10):
- $3\ \ 2 = {^5C_3} \times {^{10}C_2} = 450$
- $4\ \ 1 = {^5C_4} \times {^{10}C_1} = 50$
- $5\ \ 0 = {^5C_5} \times {^{10}C_0} = 1$
- Total $= 501$ ways
| M1, M1, A1 (3) | Mult 2 combs $^5C_x \times {^{10}C_y}$; summing 2 or 3 two-factor options, $x+y=5$; correct answer
**(iv)**
(Couple) M(4) F(9):
- ManWife $+ 3$: $0 = {^4C_3} \times {^9C_0} = 4$
- ManWife $+ 2$: $1 = {^4C_2} \times {^9C_1} = 54$
- Total $= 58$
| M1, M1, A1 (3) | Mult 2 combs $^4C_x$ and $^9C_y$; summing both options $x+y=3$, gender correct; correct answer
---6 Hannah chooses 5 singers from 15 applicants to appear in a concert. She lists the 5 singers in the order in which they will perform.\\
(i) How many different lists can Hannah make?
Of the 15 applicants, 10 are female and 5 are male.\\
(ii) Find the number of lists in which the first performer is male, the second is female, the third is male, the fourth is female and the fifth is male.
Hannah's friend Ami would like the group of 5 performers to include more males than females. The order in which they perform is no longer relevant.\\
(iii) Find the number of different selections of 5 performers with more males than females.\\
(iv) Two of the applicants are Mr and Mrs Blake. Find the number of different selections that include Mr and Mrs Blake and also fulfil Ami's requirement.
\hfill \mbox{\textit{CAIE S1 2016 Q6 [10]}}