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I assume this is a rather simple result, but I am not sure how to arrive at it.

Apparently:

$$\sum_{n=0}^6 \binom{6}{n} = 2^6$$

I can sum over all the binomial coefficients and verify this of course, but how would i arrive at the above without actually having to add all the binomial coefficients on their own?

Naz
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2 Answers2

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This follows from a cool application of the binomial formula:

$$(x+y)^m = \sum_{n=0}^m \binom{m}{n}x^ny^{m-n}.$$ I let $x=y=1$, and I let $m=6$ to get: $$2^6 = (1+1)^6 = \sum_{n=0}^6 \binom{6}{n}1^n1^{6-n} = \sum_{n=0}^6 \binom{6}{n}.$$

Mankind
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    So this reasoning can be extended to any $\sum_{n=0}^m \binom{m}{n}$. This is cool – Naz Dec 30 '15 at 15:11
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    @isquared-KeepitReal you're right! By the same method, $2^m = \sum_{n=0}^m \binom{m}{n}$. – Mankind Dec 30 '15 at 15:35
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There are $2^6$ different binary sequences of lenght $6$ (for example $111000$).

There are ${6 \choose 0}$ sequences that have no $1$s.

There are ${6 \choose 1}$ sequences that have exactly one $1$.

Continuing this we see that $$ 2^6 = \sum_{n=0}^6 {6 \choose n}\,. $$

desos
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