finding coefficients 
*

*What is the coefficient of $x^{13}y^{10}$ in $(x+y)^{23}$
Answer: $\binom{23}{10}$ or $\binom{23}{13}$ = 1,144,066


*What is the coefficient of $x^4y^6$ in $(2x + 3y)^{10}$?


Answer: $\binom{10}{6}$ or $\binom{10}{4}$ = 210
I feel like this problems are pretty simple but I just want to make sure that I have the right understanding of them. If my answers are wrong can someone explain why?
Any insight will be much appreciated!
 A: Your first answer is correct.

What is the coefficient of $x^4y^6$ in $(2x + 3y)^{10}$?

The Binomial Theorem states that
$$(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k}a^{n - k}b^k$$
In this case, $a = 2x$ and $b = 3y$.  We want the coefficient of the term $x^4y^6$ in the expansion of $(2x + 3y)^{10}$, so $n = 10$ and $k = 6$.  Therefore, the term in question is 
$$\binom{10}{6}(2x)^4(3y)^6 = 210(16x^4)(729y^6) = 2449440x^4y^6$$ 
which means the desired coefficient is $2449440$.
A: It might help to look at a simpler example. Let's take $(x+y)^3$ and look at the coefficient in front of $x^2y$. You know that the coefficient is $\binom{3}{2}=3$. This is because $(x+y)^3=(x+y)\times(x+y)\times(x+y)$ and that means we can get $x^2y$ by choosing two of the $(x+y)$'s go contribute an $x$ and one to contribute a $y$.
Now we can look at something like your second problem - let's do $(2x+3y)^3$. If you expand that out the same way, you get $(2x+3y)\times(2x+3y)\times(2x+3y)$. As before, we can get $x^2y$ by choosing two of the $(2x+3y)$'s to contribute an $x$ and one to contribute a $y$ - however, here, when we do that, the terms contribute $2x$ and $3y$ instead of just $x$ and $y$. So, each time that happens, you get $2x\times2x\times3y=12x^2y$. Since this happens $\binom{3}{2}=3$ times, overall coefficent is $\binom{3}{2}\times2^2\times3=36$.
