Question

840. n! = 7! slove the permutation
​

Answer 1

Question

Solve the permutation(Factorial over natural numbers).

Things to know

• Factorial
• One-to-one Correspondence

→ One value in the range is assigned to one value in the domain.

Solution

Factorial

$(n+1)!=(n+1)\times n!$

Strictly increasing Function

$\dfrac{(n+1)!}{n!} =n+1> 1$ hence $(n+1)!>n!$.

$n!$ is strictly increasing function over natural numbers.

So, $n!=7!\Leftrightarrow n=7$. [1]

More information

[1] If a function is strictly increasing, only one function value exists for one value(One-to-one Correspondence).

Interesting Facts

The graph of $y=x!$ is drawn by the gamma function $\Gamma(x+1)=x!$.

Gamma function was made to provide factorial values outside natural numbers, made by Euler.

The gamma function is based on these conditions.

1. $(x+1)!=(x+1)\times x!$
2. The graph must provide values outside natural numbers.

For example, we have $\Gamma(\dfrac{1}{2} )=(-\dfrac{1}{2} )!=\sqrt{\pi }$. Then, $(\dfrac{1}{2} )!=\dfrac{1}{2} \times(-\dfrac{1}{2} )!=\dfrac{\sqrt{\pi } }{2}$.

Answer 2

Answer:

Question

Solve the permutation(Factorial over natural numbers).

Things to know

Factorial

One-to-one Correspondence

→ One value in the range is assigned to one value in the domain.

Solution

Factorial

(n+1)!=(n+1)\times n!(n+1)!=(n+1)×n!

Strictly increasing Function

\dfrac{(n+1)!}{n!} =n+1 > 1

n!

(n+1)!

=n+1>1 hence (n+1)! > n!(n+1)!>n! .

n!n! is strictly increasing function over natural numbers.

So, n!=7!\Leftrightarrow n=7n!=7!⇔n=7 . [1]

More information

[1] If a function is strictly increasing, only one function value exists for one value(One-to-one Correspondence).

Interesting Facts

The graph of y=x!y=x! is drawn by the gamma function \Gamma(x+1)=x!Γ(x+1)=x! .

Gamma function was made to provide factorial values outside natural numbers, made by Euler.

The gamma function is based on these conditions.

(x+1)!=(x+1)\times x!(x+1)!=(x+1)×x!

The graph must provide values outside natural numbers.

For example, we have \Gamma(\dfrac{1}{2} )=(-\dfrac{1}{2} )!=\sqrt{\pi }Γ(

2

1

)=(−

2

1

)!=

π

. Then, (\dfrac{1}{2} )!=\dfrac{1}{2} \times(-\dfrac{1}{2} )!=\dfrac{\sqrt{\pi } }{2}(

2

1

)!=

2

1

×(−

2

1

)!=

2

π

hope it helps .

.