Prove $\left(\begin{matrix}n\\1\end{matrix}\right)=n$ - Help Needed

[bergausstein]

please help me with this

prove that

$\left(\begin{matrix}n\\1\end{matrix}\right)=n$

this is what I get when I try to prove it,

$\left(\begin{matrix}n\\1\end{matrix}\right)=\frac{n!}{1!(n-1)!}=\frac{n!}{(n-1)!}$

thanks!

 

[Nono713]

Re: Proving a corrolary

please help me with this

prove that

$\left(\begin{matrix}n\\1\end{matrix}\right)=n$

this is what I get when I try to prove it,

$\left(\begin{matrix}n\\1\end{matrix}\right)=\frac{n!}{1!(n-1)!}=\frac{n!}{(n-1)!}$

thanks!

$$n! = 2 \cdot 3 \cdot 4 \cdot \cdots \cdot (n - 1) \cdot n$$

$$(n - 1)! = 2 \cdot 3 \cdot 4 \cdot \cdots \cdot (n - 1)$$

What's left when you divide $n!$ by $(n - 1)!$? In other words, what's $n!$ in terms of $(n - 1)!$? (hint: recursive definition of factorial). If this is still unsatisfactory, consider proving by induction...

 

[bergausstein]

Re: Proving a corrolary

$$n! = 2 \cdot 3 \cdot 4 \cdot \cdots \cdot (n - 1) \cdot n$$

$$(n - 1)! = 2 \cdot 3 \cdot 4 \cdot \cdots \cdot (n - 1)$$

What's left when you divide $n!$ by $(n - 1)!$? In other words, what's $n!$ in terms of $(n - 1)!$? (hint: recursive definition of factorial). If this is still unsatisfactory, consider proving by induction...

I get your point here, but what is "recurisive definition of factorial"?

 

[Nono713]

Re: Proving a corrolary

I get your point here, but what is "recurisive definition of factorial"?

The factorial is typically defined recursively as $n! = (n - 1)! n$ and $1! = 1$, for instance:
$$5! = (5 - 1)! 5 = 4! 5 = ((4 - 1)! 4) 5 = (3! 4) 5 = ((3 - 1)! 3)4)5 = ((2! 3)4)5 = (((2 - 1)! 2)3)4)5 = (((1! 2)3)4)5 = (((2)3)4)5 = 2 \times 3 \times 4 \times 5 = 120$$
If you follow this definition, your result is immediate. If not, you can start with whatever your definition of the factorial is to arrive at the same result. For instance, if your definition is instead "product of all integers up to $n$" then you could use the product argument I gave in my previous post, or use induction (slightly tautologically, I have to say). Does that make sense?