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Using combinations find the coefficient of $x^3$ in the binomial expression of $(2x+3)^5$

I don't even know where to start, all I know is it has something to do with Pascal's triangle.

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Well, $(2x+3)^5 = (2x+3)(2x+3)(2x+3)(2x+3)(2x+3)$. Now think about how you expand such expressions. The cube terms are going to be formed by multiplying three $2x$ terms together. How many ways are there to do that? Well, we need to choose three of them, and there are 5 possibilities from which to choose. Hence, there are $\binom{5}{3}$ ways. Now when we multiply three $2x$ terms together, the resulting coefficient is going to be $2^3 = 8$. In addition, since we are multiplying over $5$ pairs of parenthesis, once we have chosen three $2x$ terms, we finish up by multiplying that by two $3$'s from the remaining two pairs of parentheses.

Hence, the final coefficient on a single $x^3$ term will be $8 \cdot 3 \cdot 3 = 72$, and there are $\binom{5}{3}$ such terms...

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Using the binomial expansion \begin{align} (t+u)^{m} = \sum_{k=0}^{m} \binom{m}{k} t^{m-k} \ u^{k} \end{align} then \begin{align} (2x+3)^{m} &= \sum_{k=0}^{m} \binom{m}{k} 2^{k} \ 3^{m-k} \ x^{k} \\ &= 3^{m} + 2 \cdot 3^{m-1} m x + 4 \cdot 3^{m-2} \binom{m}{2} x^{2} + 8 \cdot 3^{m-3} \binom{m}{3} x^{3} + \cdots \\ \end{align}

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Hint: $5\choose3$ would be the binomial coefficient of $x^3$ in $(x+1)^5$ which isn't quite what you have.

Instead of $x$ you have $2x$ and instead of $1$ you have $3$ on the end which are also worth considering here. What kind of adjustments do you think you have to make for your case?

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It is $ 5C_2*(2x)^3*3^2 $. So co-efficient would be: $10*8*9 = 720$.

This can be calculated using Binomial theorem. If you are new to the topic, you can learn quickly about Binomial theorem here.

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Perhaps you should start with a simpler problem. Pascals triangle is 1,121,1331,etc... you can find it here, but make sure you see how the next row is calculated from the previous.

Lets look at 121. What is this saying in terms of polynomials? It is on the 2nd level so it tells you it can simplify a polynomial of the form $(ax+b)^2$ for some numbers $a$ and $b$. If you are asked to simplify $(3x-2)^2$ then $a=3, b=-2$. For example, $$(ax+b)^2 = 1(ax)^2(b)^0 + 2(ax)^1(b)^1 + 1(ax)^0(b)^2$$ Do you see how the 1,2,1 were used? And how the $ax$ and $b$ were combined? Remember that anything to the power $0$ is $1$ (well, except $0$, but that's sort of an edge case)

Lets do a polynomial of degree 3. Lets do $(ax + b)^3$. Ok, where is the cheat for this in Pascals triangle? The 3rd line is 1,3,3,1. So we need: $$1(ax)^3(b)^0 + 3(ax)^2 (b)^1 + 3(ax)^1(b)^2 + 1(ax)^0(b)^3$$ So, I will show you how to get the $x^2$ term from the polynomial $(2x-3)^4$. The line in pascals triangle is 1,4,6,4,1. This means the $x^4$ term has a 1 in front, the $x^3$ term has a 4 and the $x^2$ term is then: $$6(2x)^2(-3)^2$$ Which we simplify as: $$6(4x^2)(9)=24\times 9 x^2 = 216x^2$$

Hope this helps. There are some videos at Khan Academy which explain this better.

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