Intuitive understanding of the binomial theorem? I'm trying to understand the thought process, of how I might come upon the binomial theorem intuitively by thinking about combinatorics, can someone help? 
 A: Say you want to evaluate $(a+b)^n$ then if we write this out we have
$$(a+b)^n = \underbrace{(a+b)(a+b)+\cdots+(a+b)}_{n \text{ times}}.$$
Every element in the product will be of the form $a^{k}b^{n-k}$ for $0\leq k\leq n$ if you think of how parentheses multiplication works. Now we want to say something about the coefficient in front of $a^kb^{n-k}$. When multiplying we can choose $a$ from the $k$ first parentheses and then $b$ for the remaining $n-k$. 
We can also obtain $a^{k}b^{n-k}$ by choosing $a$ for the last $k$ parentheses and the first $n-k$ we choose $b$. 
Now if you think about this you see that the coefficient is really the number of ways we can choose $k$ objects (that is $a$'s) from $n$ containers (the parentheses $(a+b)$). By basic combinatorics this number is 
$${{n}\choose{k}}.$$
Note that by choosing the parentheses we are going to take $a$ from we implicitly also make a choice of parentheses from which we will take $b$ (the remaining ones).
Therefore the coefficient of $a^{k}b^{n-k}$ is ${n}\choose{k}$ and therefore
$$(a+b)^{n} = \sum_{k=0}^{n}{{n}\choose{k}}a^{k}b^{n-k}$$
A: Maybe a example helps you: 
$(x+y)^5=(x+y)(x+y)(x+y)(x+y)(x+y)$. 
You multiply every summand with each other (distribution law). Then you can ask yourself: 
How many combinations are there where you end up with $x^5$?
There is just one combination: You take from every bracket the $x$ ($xxxxx$).
How many combinations are there where you end up with $x^4y$? There are 5 possible ways to get $x^4y$ because you could have $xxxxy$, $xxxyx$, $xxyxx$, $xyxxx$ and $yxxxx$ (you can choose one of the 5 brackets for the $y$).
How many combinations are there where you end up with $x^3y^2$? There are 10 possible ways (you choose from 5 brackets two brackets for the $y$, hence $\binom{5}{2}$.
How many combinations are there where you end up with $x^2y^3$? There are 10 possible ways ($\binom{5}{3}$). 
A: you may get an idea here
Binomial Theorem is basically expansion of terms of form $(x+y)^n$
$$(x+y)^n=\Sigma_{r=0}^n (^nC_r)\cdot x^{n-r}\cdot y^r$$  i.e. $$(x+y)^n=^nC_0\cdot x^n+^nC_1\cdot x^{n-1}y^1+....+^nC_n\cdot y^n $$ $$OR$$ $$(x+y)^n=\Sigma_{r=0}^n (^nC_r)\cdot x^r\cdot y^{n-r}$$
i.e. $$(x+y)^n=^nC_0\cdot y^n+^nC_1\cdot xy^{n-1}+...+^nC_n \cdot x^n$$
Both of these expansions are correct(why?)
For combinatorial proof: $$(x+y)^n=(x+y)\cdot(x+y)\cdot(x+y)\cdot(x+y)\cdot(x+y)..."n"times$$
Try putting small values of n and expand.
You will get to the above formula.
A: The binomial coefficient $\binom nr$ is, by definition, the number of ways to pick $r$ elements from a collection of $n$. Now take
$$
(x+y)^n=(x+y)(x+y)\cdots(x+y)
$$
and consider how many ways $x^ry^{n-r}$ can appear as you expand that product. Maybe it's easier if we add indices to the $x$'s and $y$'s, just so we can keep track of them:
$$
(x_1+y_1)(x_2+y_2)\cdots(x_n+y_n)
$$
Now, $x^ry^{n-r}$ can appear as, for instance, $x_1x_2\cdots x_ry_{r+1}\cdots y_n$. In fact, each appearance of $x^ry^{n-r}$ comes from one choice of $r$ indices from the $n$ possible indices. There are $\binom nr$ ways to do that, so $x^ry^{n-r}$ appears $\binom nr$ ways in the fully expanded product.
