Plus and Minus Makes All the Differences

Using the property that $n! = n(n-1)!$ , the expression becomes: $\frac{11(9!)}{9(9!)}\times\frac{9(7!)}{7(7!)}\times\frac{7(5!)}{5(5!)}\times\frac{5(3!)}{3(3!)}\times\ 3$ Canceling out gives $\frac{11}{9}\times\frac{9}{7}\times\frac{7}{5}\times\frac{5}{3}\times\ 3 = \boxed{11}$

10! + 9! = 10(9!) + 9! = 11(9!)

10! - 9! = 10(9!) - 9! = 9(9!)

So (10! + 9!)/(10! - 9!) = 11/9.

In a similar fashion, we have that

(8! + 7!)/(8 - 7!) = 9/7

(6! + 5!)/(6! - 5!) = 7/5

(4! + 3!)/(4! - 3!) = 5/3

(2! + 1!)/(2! - 1!) = 3/1

Multiplying all of these together gives (11/9)(9/7)(7/5)(5/3)(3/1) = 11

(10!+9!)/(10!-9!) = 9!(10+1)/9!(10-1) = (10+1)/(10-1)

Then:

(10+1)/(10-1) x (8+1)/(8-1) x (6+1)/(6-1) x (4+1)/(4-1) x (2+1)/(2-1) = 11

$n! ~=~n*\{n-1\}!\\ ~~~\dfrac{(10!+9!) (8!+7!) . . .(2!+1!)}{ (10!-9!) (8!-7!) . . .(2!-1!)}\\ =\dfrac{(10*9!+9!) (8*7!+7!) . . .(2*1!+1!)}{ (10*9!-9!) (8*7!-7!) . . .(2*1! - 1)}\\ =\dfrac{9!(10+1)~*~ 7!*(8+1) . . .1!(2+1)}{ 9!(10-1)~*~ 7!*(8-1) . . .1!(2-1) }\\ =\dfrac{9!(11)~*~ 7!*(9) . . .1!*(3)}{ 9!(9)~*~ 7!*(7) . . .1!*(1)}\\ =\dfrac{11~*~9 . . . *3}{9~*~7 . . .*1 }\\ =\dfrac{11~*~9~*~7~*~5~*~ 3}{9~*~7~*~5~*~ 3~*~1}\\ \therefore \dfrac{(10!+9!)(8!+7!)(6!+5!)(4!+3!)(2!+1!)}{(10!-9!)(8!-7!)(6!-5!)(4!-3!)(2!-1!)}\\ =\Huge \color{#D61F06}{11}.\\ ~~\\$ $~~~OR$ $\dfrac{(10!+9!)(8!+7!)(6!+5!)(4!+3!)(2!+1!)}{(10!-9!)(8!-7!)(6!-5!)(4!-3!)(2!-1!)} .\\ \implies ~\dfrac{9!(10+1)7!(8+1)5!(6+1)3!(4+1)1!(2+1)}{9!(10-1)7!(8-1)5!(6+1)3!(4+1)1! (2-1)}\\ \implies ~\dfrac{11*9*7*5*3}{9*7*5*3*1}= \color{#D61F06}{11} \\ \text{Note that 9!, 7!, 5!, 3!,1! cancel out. }$

first note that!! $\frac{(n+1)!+n!}{(n+1)! -n!} = \frac{n+2}{n}$ So given ratio simplifies to,

$\frac{11}{9} \frac{9}{7} \frac{7}{5} \frac{5}{3} \frac{3}{1} =\boxed{11}$

10! + 9! = 10(9!) + 9! = 11(9!) 10! - 9! = 10(9!) - 9! = 9(9!) Then (10! + 9!)/(10! - 9!) = 11/9 By the same way (8! + 7!)/(8 - 7!) = 9/7 (6! + 5!)/(6! - 5!) = 7/5 (4! + 3!)/(4! - 3!) = 5/3 (2! + 1!)/(2! - 1!) = 3/1 The value = (11/9)(9/7)(7/5)(5/3)(3/1) = 11

I did it a different way.

$\frac {n! + (n - 1)!}{n! - (n - 1)!} = \frac {n + 1}{n - 1}$

So the problem becomes:

$\frac {11 \times 9 \times 7 \times 5 \times 3}{9 \times 7\times 5 \times 3 \times 1}$

Most of the terms cancel out and you're left with the greatest n + 1 (ie 10 + 1 or 11).

Carried on, this sequence generates successive odd numbers.

$\frac{10!+9!}{10!-9!}=\frac{9!(10+1)}{9!(10-1)}= \frac{11}{9}$ Doing this for all factors, we get $\frac{11×9×7×5×3}{9×7×5×3×1}=\boxed{11}$

Use recursive relation: n!=(n-1)! * n

$\frac{(10!+9!)(8!+7!)(6!+5!)(4!+3!)(2!+1!)}{(10!-9!)(8!-7!)(6!-5!)(4!-3!)(2!-1!)}$

$\frac{(11*9!)(9*7!)(7*5!)(5*3!)(3*1!)}{(9*9!)(7*7!)(5*5!)(3*3!)(1*1!)}\;$

\Rightarrow \boxed{\color\red{11}}

Taking common factor of the binomials of each pair

9!(10+1) 7!(8+1) 5!(6+1) 3!(4+1) 1!(2+1)

9!(10-1) 7!(8-1) 5!(6-1) 3!(4-1) 1!(2-1)

Cancelling the binomials from up and down,remains 11

Hence, Answer = 11