Convert the following products into factorials:(i) 5 ⋅ 6 ⋅ 7

1.	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)
Answer» (i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 Now, let us evaluate Then, we can write it as: 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 = (1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10)/(1 × 2 × 3 × 4) = 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) … (2n)]/(1) (2) (3) .. (n) = (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) … (2n)] / [(2) (4) (6) … (2n)] = [(1) (2) (3) (4) … (2n-1) (2n)] / 2ⁿ [(1) (2) (3) … (n)] = (2n)! / 2ⁿ n!

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)

Answer»

(i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10

Now, let us evaluate

Then, we can write it as:

5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 = (1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10)/(1 × 2 × 3 × 4)

= 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) … (2n)]/(1) (2) (3) .. (n)

= (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) … (2n)] / [(2) (4) (6) … (2n)]

= [(1) (2) (3) (4) … (2n-1) (2n)] / 2ⁿ [(1) (2) (3) … (n)]

= (2n)! / 2ⁿ n!