Question:
Compute:
(i) $\frac{30 !}{28 !}$
(ii) $\frac{11 !-10 !}{9 !}$
(iii) L.C.M. $(6 !, 7 !, 8 !)$
Solution:
(i) $\frac{30 !}{28 !}=\frac{30 \times 29 \times 28 !}{28 !} \quad[\because n !=n(n-1) !]$
$=30 \times 29$
$=870$
(ii) $\frac{11 !-10 !}{9 !}=\frac{11 \times 10 \times 9 !-10 \times 9 !}{9 !} \quad[\because n !=n(n-1) !]$
$=\frac{9 !(110-10)}{9 !}$
$=100$
(iii) LCM of (6!,7! and 8!):
$n !=n(n-1) !$
Therefore, (6!,7! and 8!) can be rewritten as:
$8 !=8 \times 7 \times 6 !$
$7 !=7 \times 6 !$
$6 !=6 !$
$\therefore \mathrm{LCM}$ of $(6 !, 7 !$ and $8 !)=\mathrm{LCM}[8 \times 7 \times 6 !, 7 \times 6 !, 6 !]=8 \times 7 \times 6 !=8 !$
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