Question

N!/(n-4)! = 22n!/(n-3)!

Answer 1

n(n-1)(n-2)(n-3)(n-4)!        22n(n-1)(n-2)(n-3)!

-----------------------------  =  ---------------------------

             (n-4)!                                    (n-3)!

 

n-3=22

n=25

Ответ: 25

Answer 2

To solve the equation $\frac{N!}{(N-4)!} = \frac{22N!}{(N-3)!}$, we can start by simplifying both sides of the equation.

First, let's simplify the factorials:

$N! = N(N-1)(N-2)(N-3)!$ (We didn't expand the last term to keep it in a factorial form)

Now, rewrite the equation with these simplifications:

$\frac{N(N-1)(N-2)(N-3)!}{(N-4)!} = \frac{22N(N-1)(N-2)(N-3)!}{(N-3)!}$

Next, we can cancel out the common terms in the numerator and denominator on both sides:

$\frac{N(N-1)(N-2)}{(N-4)!} = \frac{22N(N-1)(N-2)}{(N-3)!}$

Now, let's consider two cases:

Case 1: $N-3 \neq 0$ (This means $N \neq 3$)

If $N-3 \neq 0$, we can cancel out $(N-3)!$ from both sides of the equation:

$\frac{N(N-1)(N-2)}{(N-4)!} = \frac{22N(N-1)(N-2)}{(N-3)!}$

Now, we have:

$\frac{N(N-1)(N-2)}{(N-4)!} = 22N(N-1)(N-2)$

Now, we can divide both sides by $N(N-1)(N-2)$ (assuming $N \neq 0, N \neq 1, N \neq 2$):

$\frac{1}{(N-4)!} = 22$

Now, we have a simple equation for $(N-4)!$:

$(N-4)! = \frac{1}{22}$

However, $(N-4)!$ must be a positive integer, but $\frac{1}{22}$ is not an integer. Therefore, there are no solutions in positive integers for this case.

Case 2: $N = 3$

If $N = 3$, then the equation becomes:

$\frac{3(3-1)(3-2)}{(3-4)!} = \frac{22 \cdot 3(3-1)(3-2)}{(3-3)!}$

Simplifying further:

$\frac{3 \cdot 2 \cdot 1}{(-1)!} = \frac{22 \cdot 3 \cdot 2 \cdot 1}{0!}$

Since $(-1)! = 1$ and $0! = 1$, both sides of the equation are equal to 6. So, the solution in this case is $N = 3$.

Therefore, the only solution in positive integers is $N = 3$.