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

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To prove that \(\left(\begin{matrix}n\\1\end{matrix}\right)=n\), the calculation begins with the formula for combinations, yielding \(\left(\begin{matrix}n\\1\end{matrix}\right)=\frac{n!}{1!(n-1)!}=\frac{n!}{(n-1)!}\). The discussion emphasizes understanding the relationship between \(n!\) and \((n-1)!\), where \(n! = n \cdot (n-1)!\). A recursive definition of factorial is introduced, stating \(n! = (n - 1)! n\) and \(1! = 1\), which simplifies the proof. The conversation suggests that using induction or the product definition of factorial can also lead to the same conclusion.
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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!
 
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Re: Proving a corrolary

bergausstein said:
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...
 
Re: Proving a corrolary

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

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

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"?
 
Re: Proving a corrolary

bergausstein said:
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?
 
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