How does $ \binom{n}{k} $ 'n choose k'
get involved with coefficient of $ (a+b)^n $. Is there any intuitive geometrical picture (interpretation) that it seems obvious?
how to visualize binomial theorem geometrically?
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$\begingroup$
combinatorics
binomial-coefficients
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1This question seems to be related: [Binomial Coefficients in the Binomial Theorem - Why Does It Work Question](http://math.stackexchange.com/questions/127926/binomial-coefficients-in-the-binomial-theorem-why-does-it-work-question) – 2012-06-11
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0Ah thanks ... !! i guess it's exactly what i'm looking for – 2012-06-11
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3Picture an $n$-dimensional cube (that is the hard part) with side length $a + b$ and divide it up... this is not easy to do beyond $n = 3$ but it is a good exercise anyway. – 2012-06-11
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0How do I divide it?? .. i mean how many smaller cubes am i going to have?? – 2012-06-11
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2There is a very good picture under "geometric explanation" in the Wikipedia article on the binomial theorem here: http://en.wikipedia.org/wiki/Binomial_theorem – 2012-06-14
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0Thanks ... you could have added an answer ... everyone would notice – 2012-06-14
3 Answers
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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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0thanks ... !! it's a lot more intuitive than comparing pascal's triangle – 2012-06-11
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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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Here is how you visualize binomial theorem https://upload.wikimedia.org/wikipedia/commons/4/47/Binomial_theorem_visualisation.svg