| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Basic committee/group selection |
| Difficulty | Moderate -0.8 This is a straightforward binomial coefficient question requiring basic algebraic manipulation. Part (a) is direct substitution into a formula (2 marks of routine work). Part (b)(i) involves expanding two binomial coefficients and simplifying to a quadratic—mechanical algebra with no conceptual difficulty. Part (b)(ii) is simply solving a quadratic equation. The entire question tests formula recall and algebraic manipulation without requiring problem-solving insight or novel approaches, making it easier than average. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.04b Binomial probabilities: link to binomial expansion |
| Answer | Marks |
|---|---|
| 7(a) | Substitutes r = 2 into given |
| Answer | Marks | Guidance |
|---|---|---|
| Condone omission of brackets | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 2.1 | R1 |
| Answer | Marks |
|---|---|
| 7(b)(i) | n |
| Answer | Marks | Guidance |
|---|---|---|
| 4! | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑙𝑙! | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ( 𝑛𝑛−2)(𝑛𝑛−3) | 2.1 | R1 |
| Answer | Marks |
|---|---|
| 7(b)(ii) | Obtains at least one correct |
| Answer | Marks | Guidance |
|---|---|---|
| PI by correct answer | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| States n = 20 | 3.2a | A1 |
| Total | 7 | |
| Q | Marking instructions | AO |
Question 7:
--- 7(a) ---
7(a) | Substitutes r = 2 into given
formula and writes out either n!
or (n-2)! correctly up to at least
three terms
Accept n! = n × (n-1) × (n-2)! or
(n-2)! = (n -2) × (n-3) × (n-4)!
Condone omission of brackets | 1.1a | M1 | nC
2 =
𝑙𝑙!
2!(𝑙𝑙−2)!
=
𝑙𝑙×(𝑙𝑙−1)×(𝑙𝑙−2)×...
2 ×(𝑙𝑙−2)×...
=
𝑙𝑙(𝑙𝑙−1)
2
Completes reasoned argument,
eliminating common factors
correctly leading to the correct
form
n×(n−1)
Must see AG
2 | 2.1 | R1
--- 7(b)(i) ---
7(b)(i) | n
Simplifies to obtain C 4 as
2n ( n−1 )( n−2 )( n−3 )
or better
4! | 1.1b | B1 | 2×nC =51×nC
4 2
=
2 𝑙𝑙! 51 𝑙𝑙(𝑙𝑙−1)
4 2 !( n 𝑙𝑙(− n 4 − ) 1 !)( n −2 )2( n−3 ) 51×n×( n−1 )
=
4! 2
(n−2)(n−3)
=51
6
n2 −5n+6=306
n2 −5n−300=0
Writes nC 4 and nC 2 in terms of n
and forms an equation using
2×nC =51×nC
4 2
Allow nC or
4
2n(n−1)(n−2)(2 n 𝑛𝑛 − ! 3)
= 4!(𝑛𝑛−4)!or
4!
nC
2 =
𝑙𝑙! | 3.1a | M1
Completes reasoned argument
to obtain given result
(n−2)(n−3)
Must see =51 or
6
= 306 OE
AG
( 𝑛𝑛−2)(𝑛𝑛−3) | 2.1 | R1
--- 7(b)(ii) ---
7(b)(ii) | Obtains at least one correct
solution to the equation
n2 −5n−300=0
PI by correct answer | 1.1a | M1 | n=−15or 20
since n > 0, so n = 20
States n = 20 | 3.2a | A1
Total | 7
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item Using ${}^n C_r = \frac{n!}{r!(n-r)!}$ show that ${}^n C_2 = \frac{n(n-1)}{2}$
[2 marks]
\item
\begin{enumerate}[label=(\roman*)]
\item Show that the equation
$$2 \times {}^n C_4 = 51 \times {}^n C_2$$
simplifies to
$$n^2 - 5n - 300 = 0$$
[3 marks]
\item Hence, solve the equation
$$2 \times {}^n C_4 = 51 \times {}^n C_2$$
[2 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2020 Q7 [7]}}