Compute - -i- 30-28- -ii- 11-10-9- -iii- L-C-M- -6-7-8- -

(i) We have 30!/28! = 30 × 29 × 28!/28! = 30 × 29 = 870 Hence, 30!/28!= 870 (ii) We have, 11! 10!/9!= 11× 10× 91 10 × 9!/9! = 9!× 10 [11 1]/9! = 10 × 10 = 10 ...

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Byju's Answer

Standard XII

Mathematics

Factorial

Compute : (i)...

Question

Compute :
(i) $\frac{30!}{28!}$ (ii) $\frac{11! - 10!}{9!}$
(iii) L.C.M. (6!,7!,8!)

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Solution

(i) We have
$\frac{30!}{28!} = \frac{30 \times 29 \times 28!}{28!}$
= $30 \times 29$
= 870
Hence, $\frac{30!}{28!} = 870$
(ii) We have,
$\frac{11! - 10!}{9!} = \frac{11 \times 10 \times 91 - 10 \times 9!}{9!}$
= $\frac{9! \times 10 [11 - 1]}{9!}$
= $10 \times 10$
= 100
Hence, $\frac{11! - 10!}{9!} = 100$
(iii) We have,
$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
and $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$
L.C.M. (6!,7!,8!)= 8!

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Similar questions

Q. Compute:
(i)

\frac{30!}{28!}

(ii)

\frac{11! - 10!}{9!}

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

Q. Express each of the following as a rational number with positive denominator:
(i)

\frac{- 15}{- 28}

(ii)

\frac{6}{- 9}

(iii)

\frac{- 28}{- 11}

(iv)

\frac{19}{- 7}

Q. Calculate the mode using tally marks for the following sizes of shoes sold out by a shopkeeper: 6, 7, 6, 5, 9, 11, 10, 7, 6, 9, 11, 9, 11, 11, 10, 9, 8, 11, 8, 9, 11, 9.

Shoe Size	Tally Marks	Frequency
5
6
7
8
9
10
11

Solve the following:

(i) x + 3 = 11; (ii) y − 9 = 21; (iii) 10 = z + 3; (iv) ; (v) 10x = 30; (vi) ;

(vii) ; (viii) ; (ix) 8x − 8 = 48; (x) ; (xi) ; (xii) ;

(xiii) 3 (x + 6) = 24; (xiv) ; (xv) 3(x + 2) − 2(x − 1) = 7.

Q. Simplify:

(i)(−8) × 9 + (−8) × 7
(ii) 9 × (−13) + 9 × (−7)
(iii) 20 × (−16) + 20 × 14
(iv) (−16) × (−15) + (−16) × (−5)
(v) (−11) × (−15) + (−11) × (−25)
(vi) 10 × (−12) + 5 × (−12)
(vii) (−16) × (−8) + (−4) × (−8)
(viii) (−26) × 72 + (−26) × 28

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Compute : (i) 30!28! (ii) 11!−10!9! (iii) L.C.M. (6!,7!,8!)

(i) We have 30!28!=30×29×28!28! = 30×29 = 870 Hence, 30!28!=870 (ii) We have, 11!−10!9!=11×10×91−10×9!9! = 9!×10[11−1]9! = 10×10 = 100 Hence, 11!−10!9!=100 (iii) We have, 8!=8×7×6×5×4×3×2×17!=7×6×5×4×3×2×1 and 6!=6×5×4×3×2×1 L.C.M. (6!,7!,8!)= 8!

Compute :
(i) $\frac{30!}{28!}$ (ii) $\frac{11! - 10!}{9!}$
(iii) L.C.M. (6!,7!,8!)