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?

 

[CellCoree]

First, we need to understand what the notation $\left(\begin{matrix}n\\1\end{matrix}\right)$ means. This is known as a binomial coefficient and can be read as "n choose 1". It represents the number of ways to choose 1 object from a set of n objects.

To prove that $\left(\begin{matrix}n\\1\end{matrix}\right)=n$, we can use the definition of a binomial coefficient and the fundamental principle of counting.

According to the fundamental principle of counting, if we have two tasks that can be performed in m and n ways respectively, then the total number of ways to perform both tasks is m x n.

In this case, we have a set of n objects and we want to choose 1 object from it. This can be seen as two tasks - first, selecting the 1 object and second, choosing the remaining (n-1) objects from the set.

The first task can be performed in n ways (as there are n objects to choose from). For the second task, we have (n-1) objects left and we want to choose (n-1) objects from it. This can be done in $\left(\begin{matrix}n-1\\n-1\end{matrix}\right)=1$ way.

Using the fundamental principle of counting, the total number of ways to choose 1 object from a set of n objects is n x 1 = n, which is equal to $\left(\begin{matrix}n\\1\end{matrix}\right)$.

Therefore, we have proven that $\left(\begin{matrix}n\\1\end{matrix}\right)=n$ for all n, and this is true by definition of a binomial coefficient.