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)

Answer 1

(i) 5.6.7.8.9.10
We have,
$5\times 6\times 7\times 8\times 9\times 10=\frac{10\times 9\times 8\times 7\times 6\times 5\times (4\times 3\times 2\times 1)}{4\times 3\times 2\times 1}=\frac{10!}{4!}$
Hence, $5\times 6\times 7\times 8\times 9\times 10=\frac{10!}{4!}$
(ii) 3.6.9.12.15.18
We have,
$3\times 6\times 9\times 12\times 15\times 18=3\times (3\times 2)\times (3\times 3)\times (3\times 4)\times (3\times 5)\times (3\times 6)$
= $3^6\times [2\times 3\times 3\times 4\times 5\times 6]=3^6\times (6!)$
(iii) (n+1)(n+2)(n+3)...(2n)
= $[1\times 2\times 3\times \mathrm{4..........}(n-1\left)n\right]\times (n+1)(n+2)...(2n-1)\times 2n$
$[1\times 2\times 3\times 4....(n-1\left)n\right]$
=$\frac{(2n!)!}{n!}$
(iv) 1.3.5.7.9 .... (2n-1)
We have,
= $1\times 3\times 5\times 7\times 9....(2n-1)$
= $\frac{\left[\mathrm{1.3.5.7.9....}\right(2n-1\left)\right].\left[\mathrm{2.4.6.8....}\right(2n-2\left)\right(2n\left)\right]}{\mathrm{2.4.6.8....}(2n-2)\left(2n\right)}$
= $\frac{\left[\mathrm{1.3.5.7.9....}\right(2n-1\left)\right].\left[\mathrm{2.4.6.8....}\right(2n-2\left)\right(2n\left)\right]}{2^n\left[\mathrm{1.2.3.4....}\right((n-1)\left(n\right)\left)\right]}$
= $\frac{\mathrm{1.2.3.4.5.6.7.8....}(2n-2)(2n-1)\left(2n\right)}{2^n,n!}=\frac{\left(2n\right)!}{2^{n}n!}$