Convert the following products into factorials:

(i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10

(ii) 3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18

(iii) (n + 1) (n + 2) (n + 3) …(2n)

(iv) 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1)

Asked by Aaryan | 1 year ago |  41

##### Solution :-

(i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10

Let us evaluate

We can write it as:

5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10

= $$\dfrac{ (1×2×3×4×5×6×7×8×9×10)}{(1×2×3×4)}$$

= $$\dfrac{ 10!}{4!}$$

(ii) 3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18

Let us evaluate

3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18 = (3×1) × (3×2) × (3×3) × (3×4) × (3×5) × (3×6)

= 36 (1×2×3×4×5×6)

= 36 (6!)

(iii) (n + 1) (n + 2) (n + 3) … (2n)

Let us evaluate

(n + 1) (n + 2) (n + 3) … (2n) = [(1) (2) (3) .. (n) … (n + 1) (n + 2) (n + 3) …

$$\dfrac{ (2n) }{ (1) (2) (3) .. (n)}$$

= $$\dfrac{ (2n)!}{n!}$$

(iv) 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1)

Let us evaluate

1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1) = [(1) (3) (5) … (2n-1)] [(2) (4) (6) …

$$\dfrac{ (2n) }{ (2) (4) (6) … (2n)}$$

= [(1) (2) (3) (4) … $$\dfrac{ (2n-1) (2n)] }{2n [(1) (2) (3) … (n)]}$$

=$$\dfrac{ (2n)! }{ 2^n n!}$$

Answered by Aaryan | 1 year ago

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