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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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This question seems to be related: Binomial Coefficients in the Binomial Theorem - Why Does It Work Question – Martin Sleziak Jun 11 '12 at 13:03
Ah thanks ... !! i guess it's exactly what i'm looking for – Santosh Linkha Jun 11 '12 at 13:04
Picture 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. – Qiaochu Yuan Jun 11 '12 at 14:07
How do I divide it?? .. i mean how many smaller cubes am i going to have?? – Santosh Linkha Jun 11 '12 at 17:01
There is a very good picture under "geometric explanation" in the Wikipedia article on the binomial theorem here: – Jair Taylor Jun 14 '12 at 19:01
up vote 5 down vote accepted

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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thanks ... !! it's a lot more intuitive than comparing pascal's triangle – Santosh Linkha Jun 11 '12 at 13:07

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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