I am just trying to understand binomial coefficients (nCr) from a combinatorial point of view, instead of the Pascal triangle.
Consider (x + y)^n, where the coefficient of the term x^(n-r)y^r is nCr.
My intuition is that say we throw 2 coins each time, and do this for n times (the intuition being that the first coin represents the variable x and the second coin represents the variable y).
And note the number of times where the first/second coin shows Head/Tail, with the condition being that after throwing this pair of coins for n times, the total number of Head/Tail is always equal to n (the intuition being that the power of x^(n-r) and the power of y^r always add up to (n-r) + r = n).
I haven't managed to show this mathematically. Hope my intuition is correct and that someone can help. :)
To see why the coefficient of $x^{n-r}y^r$ in $(x+y)^n$ is $n \choose r$, imagine writing each copy of $(x+y)$ in the expansion $$(x+y)^n=(x+y)(x+y)\cdots(x+y)$$ in a different color ink. Then the coefficient of $x^{n-r}y^r$ is simply the number of ways one can pick $n-r$ colors out of $n$ colors. Each such choice corresponds with one subset of size $n-r$ of the $n$-set of copies of $(x+y)$, that subset being the copies contributing the $n-r$ factors of $x$ that go into $x^{n-r}y^r$. Of course, picking any $n-r$ colors to contribute the $x$’s leaves the remaining $r$ colors as the ones contributing the $y$’s.
With coin tosses, you could ask, “How many sequences of $n$ tosses of a single coin include exactly $n-r$ heads (and therefore also exactly $r$ tails)?” The answer amounts to the number of ways in which one can pick $n-r$ places out of $n$ to fill with the symbol H, with the remaining $r$ places filled with T’s.