| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Committee with gender/category constraints |
| Difficulty | Moderate -0.3 This is a straightforward combinations question with standard constraints. Part (i) is direct application of C(26,8), part (ii) requires multiplying two combinations C(12,4)×C(14,4), and part (iii) needs casework (5,6,7,8 adults) but follows routine patterns. While it's a multi-part question requiring careful organization, it involves only standard combinatorial techniques with no novel insight or complex reasoning required. Slightly easier than average due to its mechanical nature. |
| Spec | 5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(^{26}C_8 = 1562275\) | B1 | Correct answer 1562275 (no need to see calculation) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(^{12}C_4 \times ^{14}C_4\) | M1 | Attempts the product shown |
| \(= 495 \times 1001 = 495495\) | A1 | 495495 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts cases for 5, 6, 7 or 8 adults on team (only) | M1 | Works out different cases giving more than half adults. Need not be correct formula; do not allow if 4 adults/4 juniors case is included |
| \(^{12}C_5 \times ^{14}C_3 + ^{12}C_6 \times ^{14}C_2 + ^{12}C_7 \times ^{14}C_1 + ^{12}C_8 \times ^{14}C_0 = \ldots\) | M1 | Sums the possibilities; binomial products must be correct; allow if 4/4 case included |
| \(= 383955\) | A1 | cao |
## Question 1:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $^{26}C_8 = 1562275$ | B1 | Correct answer 1562275 (no need to see calculation) |
**(1 mark)**
---
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $^{12}C_4 \times ^{14}C_4$ | M1 | Attempts the product shown |
| $= 495 \times 1001 = 495495$ | A1 | 495495 |
**(2 marks)**
---
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts cases for 5, 6, 7 or 8 adults on team (only) | M1 | Works out different cases giving more than half adults. Need not be correct formula; do not allow if 4 adults/4 juniors case is included |
| $^{12}C_5 \times ^{14}C_3 + ^{12}C_6 \times ^{14}C_2 + ^{12}C_7 \times ^{14}C_1 + ^{12}C_8 \times ^{14}C_0 = \ldots$ | M1 | Sums the possibilities; binomial products must be correct; allow if 4/4 case included |
| $= 383955$ | A1 | cao |
**(3 marks)**
---\begin{enumerate}
\item A small sports club has 12 adult members and 14 junior members.
\end{enumerate}
The club needs to enter a team of 8 players for a particular competition.\\
Determine the number of ways in which the team can be selected if\\
(i) there are no restrictions on the team,\\
(ii) the team must contain 4 adults and 4 juniors,\\
(iii) more than half the team must be adults.
\hfill \mbox{\textit{Edexcel FP2 2020 Q1 [6]}}