AQA Paper 3 2019 June — Question 2 1 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2019
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCombinations & Selection
TypeBasic committee/group selection
DifficultyEasy -1.8 This is a straightforward factorial simplification requiring only basic cancellation: 100!/(98!×3!) = (100×99)/(3×2×1) = 1650. It's a 1-mark multiple choice question testing routine manipulation of factorials with no problem-solving required, making it significantly easier than average.
Spec5.01a Permutations and combinations: evaluate probabilities

Find the value of \(\frac{100!}{98! \times 3!}\) Circle your answer. [1 mark] \(\frac{50}{147}\) \quad \(1650\) \quad \(3300\) \quad \(161700\)

Question 2:
AnswerMarks Guidance
2Circles the correct response 1.1b
Total1
QMarking instructions AO
Question 2:
2 | Circles the correct response | 1.1b | B1 | 1650
Total | 1
Q | Marking instructions | AO | Mark | Typical solution
Find the value of $\frac{100!}{98! \times 3!}$

Circle your answer.
[1 mark]

$\frac{50}{147}$ \quad $1650$ \quad $3300$ \quad $161700$

\hfill \mbox{\textit{AQA Paper 3 2019 Q2 [1]}}