| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 4 |
| Topic | Combinations & Selection |
| Type | Basic committee/group selection |
| Difficulty | Moderate -0.3 Part (i) is a straightforward algebraic manipulation of the binomial coefficient formula requiring only basic factorial cancellation. Part (ii) involves substituting the result from (i) and solving a quadratic equation in n, which is routine algebra. The question tests standard binomial coefficient manipulation with no novel insight required, making it slightly easier than average but not trivial due to the multi-step nature of part (ii). |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.04b Binomial probabilities: link to binomial expansion |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\binom{n}{n-2} = \frac{n!}{(n-2)!2!} = \frac{n(n-1)}{2}\) | B1, (AG) | 1 mark |
| (ii) Obtain \(\frac{(2n+1)2n}{2} - 2 \times \frac{n(n-1)}{2} = 24\) | M1 | Simplify to \(n^2 + 2n - 24 = 0\) |
(i) $\binom{n}{n-2} = \frac{n!}{(n-2)!2!} = \frac{n(n-1)}{2}$ | B1, (AG) | 1 mark
(ii) Obtain $\frac{(2n+1)2n}{2} - 2 \times \frac{n(n-1)}{2} = 24$ | M1 | Simplify to $n^2 + 2n - 24 = 0$ | A1 | Obtain $n = 4$ only | A1 | 3 marks\begin{enumerate}[label=(\roman*)]
\item Show that $\binom{n}{n-2} = \frac{n(n-1)}{2}$, where the positive integer $n$ satisfies $n \geqslant 2$. [1]
\item Solve the equation $\binom{2n+1}{2n-1} - 2 \times \binom{n}{n-2} = 24$. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q1 [4]}}