How exactly is Newton's Binomial Theorem Proven Using Combinatorial Ideas?

Question

Let $n$ and $k$ be nonnegative integers. The binomial coefficient $\binom{n}{k}$ is equal to the number of ways to choose $k$ items from a set with $n$ elements. How can this interpretation of $\binom{n}{k}$ be used to understand or prove the binomial theorem $$ (x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k} \qquad \text{for all numbers } x, y. $$

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Note: The original text of this question is shown below. I’m editing the question because it’s fundamental and deserves to be salvaged.

Original text:

Sorry for the childish question. I am in 7th grade and need some help. Thank You!

Answer 1

Let's imagine what happens if we expand $$ (x+y)^3 = \underbrace{(x+y)}_{\text{factor 1}} \underbrace{(x+y)}_{\text{factor 2}} \underbrace{(x+y)}_{\text{factor 3}} $$ without collecting like terms. The result is \begin{align} (x + y)^3 &= x\cdot x \cdot x \\ & + x \cdot x \cdot y + x \cdot y \cdot x + y \cdot x \cdot x \\ &+ x \cdot y \cdot y + y \cdot x \cdot y + y \cdot y \cdot x \\ &+ y \cdot y \cdot y. \end{align} Each term in the sum contains a contribution from factor 1, a contribution from factor 2, and a contribution from factor 3. Each factor contributes either an $x$ or a $y$.

Now how many terms in this sum are equal to $x y^2$? Well, how many ways are there to choose two of the factors to contribute a $y$? The answer is $\binom{3}{2}$.

Likewise, the sum contains $\binom{3}{0}$ terms that are equal to $x^3$, $\binom{3}{1}$ terms that are equal to $x^2 y$, and $\binom{3}{3}$ terms that are equal to $y^3$.

That explains why $$ (x + y)^3 = \binom{3}{0} x^3 + \binom{3}{1}x^2 y + \binom{3}{2} x y^2 + \binom{3}{3} y^3. $$

Answer 2

HINT:

The Binomial Theorem says: $$ (a+b)^n=\sum_{k=0}^{n}\binom{n}{k} a^{n-k}b^k, $$ where the left hand side is $$ \underbrace{(a+b)\cdot(a+b)\cdot(a+b)\cdot\ldots\cdot(a+b)}_{n \ times}. $$ If you would expand this product you would pick one element from each bracket. This last sentence could already ring some bells since combinatorics usually deals with questions "In now many different ways can I pick...".

For a more detailed explanation you could look at the wikipedia page.

Hope this helped a bit and you are always welcome to add further questions and ideas as a comment on my answer