A little difference

Relevant wiki: Factorials Problem Solving - Basic

$\begin{aligned} P & = \frac {(6!+5!) \times (4!+3!) \times (2!+1!)}{(6!-5!) \times (4!-3!) \times (2!-1!)} & \small \color{#3D99F6} \text{Note that }n! = n(n-1)(n-2)\cdots \times 3\times 2\times 1 \\ & = \frac {(6\times 5!+5!) \times (4 \times 3!+3!) \times (2 \times 1!+1!)}{(6\times 5!-5!) \times (4\times 3!-3!) \times (2\times 1!-1!)} & \small \color{#3D99F6} \implies n! = n(n-1)! \\ & = \frac {\cancel{5!}(6+1)}{\cancel{5!}(6-1)} \times \frac {\cancel{3!}(4+1)}{\cancel{3!}(4-1)} \times \frac {\cancel{1!}(2+1)}{\cancel{1!}(2-1)} \\ & = \frac 7{\cancel{5}} \times \frac {\cancel{5}}{\cancel{3}} \times \frac {\cancel{3}}1 \\ & = \boxed{7} \end{aligned}$

Relevant wiki: Factorials Problem Solving - Basic

$\dfrac{(6! + 5!) (4! + 3!) (2! + 1!)}{(6! - 5!) (4! - 3!) (2! - 1!)} = \dfrac{5!(6+1) \times 3!(4+1) \times 1!(2+1)}{5!(6-1) \times 3!(4-1) \times 1!(2-1)} = \dfrac{7 \times 5 \times 3}{5 \times 3} = \boxed{7}$

$\dfrac{(6! + 5!) \times (4! + 3!) \times (2! + 1!)}{(6! - 5!) \times (4! - 3!) \times (2! - 1!)} =$ $\dfrac{5! (7) \times 3! (5) \times 1! (3)}{5! (5) \times 3! (3) \times 1! (1)} =$ $\dfrac{7 \times 5 \times 3}{5 \times 3 \times 1} = \boxed{7}.$

In general, $(n+1)! + n! = (n+2)\cdot n!$ and $(n+1)! - n! = n\cdot n!$ .

Thus, $6! + 5! = 7\cdot 5!$ and $6! - 5! = 5\cdot 5!$ . The fraction simplifies to $\frac{7\cdot 5! \times 5\cdot 3! \times 3\times 1!}{5\cdot 5! \times 3\cdot 3! \times 1\times 1!} = \frac{7\times 5\times 3}{5\times 3\times 1} = \boxed{7}.$

1. Multiply all of the top terms together.
2. Add them.
3. Multiply all of the bottom terms.
4. Add them.
5. Divide the numerator by the denominator to get 7.

I tried out just the first fraction or part of the expression and saw the answer was 7/5. So then i thought the pattern repeats. The biggest number that's a factorial gets one added to it and turns into a numerator of a new fraction. The denominator of that fraction is the smaller number with the factorial. So it's 7/5 * 5/3 * 3/1, which is 7.

$\frac{(6!+5!)(4!+3!)(2!+1!)}{(6!-5!)(4!-3!)(2!-1!)}$ = $\frac{5!(6+1)3!(4+1)(2+1)}{5!(6-1)3!(4-1)(2-1)}$ = $\frac{7·5·3}{5·3·1}$ =7

Slow solution: (840×30×3)/(600×18×1)

O.K. this took me a while, but here is how I went through it: $(6!-5!)\times(4!-3!)=30(4!)^{2}-120(3!)^{2}-5(4!)^{2}+20(3!)^{2}$ if you foil it. Which then simplifies to: $25(4!)^{2}-100(3!)^{2}.$ I did the same thing with the numerator but since all the operations in the parenthesis are addition it is the same except with all plus signs: $30(4!)^{2}+120(3!)^{2}+5(4!)^{2}+20(3!)^{2}.$ That then simplifies to $35(4!)^{2}+140(3!)^{2}.$ Then you can quickly say that $(2!-1!)$ is easily 1, and $(2!+1!)$ is 3. So the fraction is now $\dfrac{(35(4!)^{2}+140(3!)^{2}\times3}{(25(4!)^{2}-100(3!)^{2}}.$ Then by painstakingly multiplying I found out that $(4!)^{2}$ is 576, and $(3!)^{2}$ is 36. So now the fraction is $\dfrac{[35(576)+140(36)]\times3}{25(576)-100(36)}.$ This equals $\dfrac{(20160+5040)3}{14400-3600}$ = $\dfrac{25200\times3}{10800}$ which equals $\dfrac{75600}{10800}$ and that is 7.

\frac {(n+1)! + n!}{(n+1)!-n!} = \frac {n+2}{n}