CAIE IGCSE Additional Math 11 Permutations and Combinations Topic Practice

$$\frac{(n-4)(n+1)!}{(n-4)!5!}=\frac{(n+2)!}{(n-5)!7!}$$

leading to
$$\times 6=n+2$$ B1 for $7 \times 6$ or 42 seen as a result of simplification of the numerical factorials

B1 for n+2 seen as a result of simplification of the algebraic factorials
n=40 B1

A correct and simplified equation e.g.
$$n+1=\frac{437}{n-3}$$
B1 for an equation with
$$\left[\frac{(n+1)!}{n!}=\right] n+1 \text { and } 437$$
or $\left[\frac{(n-3)!}{(n-4)!}\right.$ oe $\left.=\right] n-3$ and 437
OR
for $\frac{(n+1)!}{437(n-4)!}=\frac{n!}{(n-3)!}$$n^{2}-2 n-440[=0]$
dep on B2

n=22 as only solution
dep on all previous marks awarded