Compute:

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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