Question

Compute:
(i) $\frac{30!}{28!}$

(ii) $\frac{11!-10!}{9!}$

(iii) L.C.M. (6!, 7!, 8!)

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\frac{30!}{28!}=\frac{30\times 29\times 28!}{28!}\left[\because n!=n(n-1)!\right] \\ =30\times 29 \\ =870 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\frac{11!-10!}{9!}=\frac{11\times 10\times 9!-10\times 9!}{9!}\left[\because n!=n(n-1)!\right] \\ =\frac{9!(110-10)}{9!} \\ =100 \end{gathered}$$

(iii) LCM of (6!,7! and 8!):
n! = n(n$-$1)!
Therefore, (6!,7! and 8!) can be rewritten as:
8! = 8$\times$7$\times$6!
7! = 7$\times$6!
6! = 6!
∴ LCM of (6!,7! and 8!) = LCM [8$\times$7$\times$6!, 7$\times$6! , 6!] = 8$\times$7$\times$6! = 8!