# Convert the following products into factorials:

Question:

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)

Solution:

(i) $5 \times 6 \times 7 \times 8 \times 9 \times 10=\frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10}{1 \times 2 \times 3 \times 4}$

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

(ii) $3 \times 6 \times 9 \times 12 \times 15 \times 18=(3 \times 1) \times(3 \times 2) \times(3 \times 3) \times(3 \times 4) \times(3 \times 5) \times(3 \times 6)$

$=3^{6}(1 \times 2 \times 3 \times 4 \times 5 \times 6)$

$=3^{6}(6 !)$

(iii) $(n+1)(n+2)(n+3) \ldots(2 n)=\frac{(1)(2)(3) \ldots(n)(n+1)(n+2)(n+3) \ldots(2 n)}{(1)(2)(3) \ldots(\mathrm{n})}$

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

(iv) (1) (3) (5)\ldots....... (2n - 1) $=\frac{[(1)(3)(5) \ldots \ldots \ldots(2 n-1)][(2)(4)(6) \ldots \ldots \ldots .(2 n)]}{[(2)(4)(6) \ldots \ldots \ldots(2 n)]}$

$=\frac{[(1)(2)(3)(4)(5) \ldots \ldots \ldots .(2 n-1)(2 n)]}{2^{n}[(1)(2)(3) \ldots \ldots \ldots(n)]}$

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