1. ## Binomial coefficients identities

Hey.

I am struggling with the formula for binomial coefficients.

$(x+y)^n=\binom{n}{0}x^ny^0+...\binom{n}{n}x^0y^n$

I understand intuitively why x and y has these increasing and decreasing exponents, but I don't catch the method of finding the suitable coefficients by the way of "choose".

Some clarification would really help me.

Thanks.

In this context, I am trying to prove these attached identities for every Natural n >= 2.

3. ## Re: Binomial coefficients identities

In the use of the binomial coefficients in the binomial theorem, consider if we have $\displaystyle n$ factors:

$\displaystyle y=(x+y)^n=(x+y)\cdots(x+y)$

Now, if we are curious how many ways we can get $\displaystyle x^m$ where $\displaystyle m\le n$ we see that we are choosing $\displaystyle m$ $\displaystyle x$'s from a total of $\displaystyle n$ $\displaystyle x$'s. And so we know:

$\displaystyle {n \choose m}x^my^{n-m}$

will be one of the terms in the expansion. Does this make sense intuitively?

4. ## Re: Binomial coefficients identities

Originally Posted by CStudent
I am struggling with the formula for binomial coefficients.
$(x+y)^n=\binom{n}{0}x^ny^0+...\binom{n}{n}x^0y^n$
For the sake illustration lets say $n>5$ consider: $\displaystyle {(x + y)^n} = (x + y)(x + y)(x + y)\underbrace \cdots _{n - 5}(x + y)(x + y)$.
In going across that product we stop at term and take either $x\text{ or } y$ When done we have $k~x's~\&~j~y's$ and $k+j=n$
You have seen this idea played out many times. $xxxxxxxyyyy$ how many ways can those eleven be rearranged?
$\dfrac{11!}{7!\cdot 4!}=\dbinom{11}{7}=330$ SEE HERE
That is like stopping eleven times picking $7~x's~\&~4 y's$ in random order. That can be done in 330 ways.
Thus $330x^7y^4$ is in that expansion. SEE HERE
Note the both $330x^7y^4~\&~330x^4y^7$ are terms in the expansion. Why is that?

In the expansion of $(x+y)^{21}$, what is the coefficient of the $x^{10}y^{11}$ term?

5. ## Re: Binomial coefficients identities

Originally Posted by Plato
For the sake illustration lets say $n>5$ consider: $\displaystyle {(x + y)^n} = (x + y)(x + y)(x + y)\underbrace \cdots _{n - 5}(x + y)(x + y)$.
In going across that product we stop at term and take either $x\text{ or } y$ When done we have $k~x's~\&~j~y's$ and $k+j=n$
You have seen this idea played out many times. $xxxxxxxyyyy$ how many ways can those eleven be rearranged?
$\dfrac{11!}{7!\cdot 4!}=\dbinom{11}{7}=330$ SEE HERE
That is like stopping eleven times picking $7~x's~\&~4 y's$ in random order. That can be done in 330 ways.
Thus $330x^7y^4$ is in that expansion. SEE HERE
Note the both $330x^7y^4~\&~330x^4y^7$ are terms in the expansion. Why is that?

In the expansion of $(x+y)^{21}$, what is the coefficient of the $x^{10}y^{11}$ term?
I almost understand everything here but not absolutely.
How do we know that in the equation we actually choose x's and y's the way it is determined with the choose?
I don't get it to end, how are you sure we "compose" x's and y's in equation according to that choose and not simply let's say factorial of 11?

Thanks.

6. ## Re: Binomial coefficients identities

Originally Posted by MarkFL
In the use of the binomial coefficients in the binomial theorem, consider if we have $\displaystyle n$ factors:

$\displaystyle y=(x+y)^n=(x+y)\cdots(x+y)$

Now, if we are curious how many ways we can get $\displaystyle x^m$ where $\displaystyle m\le n$ we see that we are choosing $\displaystyle m$ $\displaystyle x$'s from a total of $\displaystyle n$ $\displaystyle x$'s. And so we know:

$\displaystyle {n \choose m}x^my^{n-m}$

will be one of the terms in the expansion. Does this make sense intuitively?
Yes and no.
I don't get it to the end.
See my reply above to Plato.
Thanks.

7. ## Re: Binomial coefficients identities

Originally Posted by romsek
That's the problem with proofs with induction.
It really proves theorems but it doesn't helps you understand the logic and the scenes behind the theorem.
Thanks.