How does $ \binom{n}{k} $ 'n choose k' get involved with coefficient of $ (a+b)^n $. Is there any intuitive geometrical picture (interpretation) that it seems obvious?
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Hint: Imagine writing $(a+b)^n$ as $(a+b)(a+b)\dots(a+b)$, and then multiplying out all the brackets. Ask yourself how many ways you can get a term involving $a^kb^{n-k}$. |
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Expanding on what Old John wrote, it might help to consider a "noncommutative" version of the binomial theorem. $(a+b)^n = (a+b)(a+b)...(a+b)$ is going to have $2^n$ terms. Each of the $2^n$ words of length $n$ consisting of the letters $a$ and $b$ will occur exactly once. If you identify words via commutativity of multiplication, you will see there are $\binom{n}{k}$ words in the equivalence class of $a^{n-k}b^k$. |
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